Probability

A box contains 600 screws, one-tenth are rusted. One screw is taken out at random from this box. Find the probability that it is a good screw.

Answer

No. of rusted screws = $\dfrac{1}{10} \times 600$ = 60,

No. of good screws = 600 - 60 = 540,

Let E₁ be the event of taking out a good screw, then number of favourable outcomes to E₁ = 540

$P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{540}{600} = \dfrac{9}{10}.$

Hence, the probability of taking out a random screw is $\dfrac{9}{10}$.

In a lottery, there are 5 prized tickets and 995 blank tickets. A person buys a lottery ticket. Find the probability of his winning a prize.

Answer

Total no. of tickets = 995 + 5 = 1000

Let E₁ be the event of getting a prized ticket, then number of favourable outcomes to E₁ = 5

$P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{5}{1000} = \dfrac{1}{200}.$

Hence, the probability of winning a prize is $\dfrac{1}{200}$.

12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Answer

No. of defective pens = 12

No. of good pens = 132

Total no. of pens = 12 + 132 = 144.

Let E₁ be the event of getting a good pen, then number of favourable outcomes to E₁ = 132

$P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{132}{144} = \dfrac{11}{12}.$

Hence, the probability of taking out a good pen is $\dfrac{11}{12}$.

Two players, Sania and Sonali, play a tennis match. It is known that the probability of Sania winning the match is 0.69. What is the probability of Sonali winning?

Answer

Let P(E) be the probability of Sania's winning and P(E') be the probability of Sania losing or the probability of Sonali winning.

∴ P(E) + P(E') = 1

⇒ 0.69 + P(E') = 1
⇒ P(E') = 1 - 0.69
⇒ P(E') = 0.31

Hence, the probability of Sonali winning is 0.31

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) red?

(ii) not red?

Answer

(i) No. of red balls = 3
    No. of black balls = 5.

Total no. of balls = 3 + 5 = 8.

Let E₁ be the event of drawing a red ball, then number of favourable outcomes to E₁ = 3

$P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{3}{8}.$

Hence, the probability of drawing out a red ball is $\dfrac{3}{8}$.

(ii) Since, there are only two coloured balls red and black. Hence, the probability of not drawing a red ball = probability of drawing a black ball.

Let E₂ be the event of drawing a black ball, then number of favourable outcomes to E₂ = 5

$P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{5}{8}.$

Hence, the probability of drawing out a not red ball is $\dfrac{5}{8}$.

A letter is chosen from the word 'TRIANGLE'. What is the probability that it is a vowel?

Answer

Vowels in 'TRIANGLE' = 'I', 'A', 'E'.

No. of letters in 'TRIANGLE' = 8.

Let E₁ be the event of choosing a vowel, then number of favourable outcomes to E₁ = 3

$P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{3}{8}.$

Hence, the probability of choosing a vowel is $\dfrac{3}{8}$.

A letter of English alphabet is choosen at random. Determine probability that the letter is consonant.

Answer

Total alphabets = 26

No. of vowels = 5

No. of consonants = 21

Let E₁ be the event of choosing a consonant, then number of favourable outcomes to E₁ = 21

$P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{21}{26}.$

Hence, the probability of choosing a consonant is $\dfrac{21}{26}$.

A bag contains 5 white, 2 red and 3 black balls. A ball is drawn at random. What is the probability that the ball drawn is red ball?

Answer

Total number of balls = 5 + 2 + 3 = 10

So, the total number of possible outcomes = 10

There are 2 red balls.

∴ Number of favourable outcomes = 2

P(getting a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{10} = \dfrac{1}{5}$.

Hence, probability of getting a red ball = $\dfrac{1}{5}$.

A box contains 7 blue, 8 white and 5 black marbles. If a marble is drawn at random from the box, what is the probability that it will be

(i) black?

(ii) blue or black?

(iii) not black?

(iv) green?

Answer

(i) No. of black marbles = 5 and Total marbles = 7 + 8 + 5 = 20.

Let E₁ be the event of choosing a black marble, then number of favourable outcomes to E₁ = 5

$P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{5}{20} = \dfrac{1}{4}.$

Hence, the probability of choosing a black marble is $\dfrac{1}{4}$.

(ii) Total no. of black and blue marbles = 7 + 5 = 12.

Let E₂ be the event of choosing a black or blue marble, then number of favourable outcomes to E₂ = 12

$P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{12}{20} = \dfrac{3}{5}.$

Hence, the probability of choosing a black or blue marble is $\dfrac{3}{5}$.

(iii) Total no. of blue and white marbles = 7 + 8 = 15.

Let E₃ be the event of choosing a white or blue marble, then number of favourable outcomes to E₃ = 15

$P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{15}{20} = \dfrac{3}{4}.$

Hence, the probability of choosing a not black marble is $\dfrac{3}{4}$.

(iv) Let E₄ be the event of choosing a green marble, then number of favourable outcomes to E₄ = 0, as there is no green marble in the box.

$P(E_4) = \dfrac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \dfrac{0}{20} = 0.$

Hence, the probability of choosing a green marble is 0.

A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls. One ball is drawn at random from the bag. Find the probability that the ball is :

(i) white

(ii) red or black

(iii) not green

(iv) neither white nor black.

Answer

(i) No. of white balls = 8 and total no. of balls = 6 + 8 + 5 + 3 = 22.

Let E₁ be the event of choosing a white ball, then number of favourable outcomes to E₁ = 8

$P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{8}{22} = \dfrac{4}{11}.$

Hence, the probability of drawing a white ball is $\dfrac{4}{11}$.

(ii) Total no. of red and black balls = 6 + 3 = 9.

Let E₂ be the event of choosing a red or black ball, then number of favourable outcomes to E₂ = 9

$P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{9}{22}.$

Hence, the probability of drawing a red or black ball is $\dfrac{9}{22}$.

(iii) Probability of not drawing a green ball means probability of drawing any other colour ball.

Total no. of red, black and white balls = 6 + 3 + 8 = 17.

Let E₃ be the event of choosing a red, black or white ball, then number of favourable outcomes to E₃ = 17

$P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{17}{22}.$

Hence, the probability of drawing a not green ball is $\dfrac{17}{22}$.

(iv) Probability of not drawing a white or black ball means probability of drawing red or green ball.

Let E₄ be the event of choosing a red or green ball, then number of favourable outcomes to E₄ = 11

$P(E_4) = \dfrac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \dfrac{11}{22} = \dfrac{1}{2}.$

Hence, the probability of drawing neither white nor black ball is $\dfrac{1}{2}$.

A carton consist of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Peter, a trader, will only accept the shirts which are good, but Salim, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that

(i) it is acceptable to Peter?

(ii) it is acceptable to Salim?

Answer

(i) No. of good shirts = 88
    Total no. of shirts = 100.

Let E₁ be the event of choosing a good shirt, then number of favourable outcomes to E₁ = 88

$P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{88}{100} = \dfrac{22}{25}.$

Hence, the probability that shirt is acceptable to Peter is $\dfrac{22}{25}$.

(ii) No. of good and minor defective shirts = 88 + 8 = 96
    Total no. of shirts = 100.

Let E₂ be the event of choosing a good or minor defective shirt, then number of favourable outcomes to E₂ = 96

$P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{96}{100} = \dfrac{24}{25}.$

Hence, the probability that shirt is acceptable to Salim is $\dfrac{24}{25}$.

A die is thrown once. What is the probability that the

(i) number is even

(ii) number is greater than 2?

Answer

(i) Dice is thrown once
    Sample space = {1, 2, 3, 4, 5, 6} which has 6 likely outcomes.

Let E₁ be the event of getting a even number, then number of favourable outcomes to E₁ = 3

$P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}.$

Hence, the probability of getting a even no. is $\dfrac{1}{2}$.

(ii) Let E₂ be the event of getting a number greater than 2, then number of favourable outcomes to E₂ = 4

$P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{4}{6} = \dfrac{2}{3}.$

Hence, the probability of getting a no. greater than 2 is $\dfrac{2}{3}$.

In a single throw of a die, find the probability of getting :

(i) an odd number

(ii) a number less than 5

(iii) a number greater than 5

(iv) a prime number

(v) a number less than 7

(vi) a number divisible by 3

(vii) a number between 3 and 6

(viii) a number divisible by 2 or 3.

Answer

In a single throw of die,

Sample space = {1, 2, 3, 4, 5, 6}.

(i) Let E be the event of getting an odd number, then

E = {1, 3, 5}.

∴ The number of favourable outcomes to the event E = 3.

$\therefore P(E) = \dfrac{\text{No. of favourable outcomes to } E}{\text{Total no. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}.$

Hence, the probability of getting an odd number is $\dfrac{1}{2}$.

(ii) Let E₁ be the event of getting a number less than 5, then

E₁ = {1, 2, 3, 4}.

∴ The number of favourable outcomes to the event E₁ = 4.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{4}{6} = \dfrac{2}{3}.$

Hence, the probability of getting a number less than 5 is $\dfrac{2}{3}$.

(iii) Let E₂ be the event of getting a number greater than 5, then

E₂ = {6}.

∴ The number of favourable outcomes to the event E₂ = 1.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{1}{6}.$

Hence, the probability of getting a number greater than 5 is $\dfrac{1}{6}$.

(iv) Let E₃ be the event of getting a prime number, then

E₃ = {2, 3, 5}.

∴ The number of favourable outcomes to the event E₃ = 3.

$\therefore P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}.$

Hence, the probability of getting a prime number is $\dfrac{1}{2}$.

(v) Let E₄ be the event of getting a number less than 7, then

E₄ = {1, 2, 3, 4, 5, 6}.

∴ The number of favourable outcomes to the event E₄ = 6.

$\therefore P(E_4) = \dfrac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \dfrac{6}{6} = 1.$

Hence, the probability of getting a number less than 7 is 1.

(vi) Let E₅ be the event of getting a number divisible by 3, then

E₅ = {3, 6}.

∴ The number of favourable outcomes to the event E₅ = 2.

$\therefore P(E_5) = \dfrac{\text{No. of favourable outcomes to } E_5}{\text{Total no. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}.$

Hence, the probability of getting a number divisible by 3 is $\dfrac{1}{3}$.

(vii) Let E₆ be the event of getting a number between 3 and 6.

E₆ = {4, 5}.

∴ The number of favourable outcomes to the event E₆ = 2.

$\therefore P(E_6) = \dfrac{\text{No. of favourable outcomes to } E_6}{\text{Total no. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}.$

Hence, the probability of getting a number between 3 and 6 is $\dfrac{1}{3}$.

(viii) Let E₇ be the event of getting a number divisible by 2 or 3, then

E₇ = {2, 3, 4, 6}.

∴ The number of favourable outcomes to the event E₇ = 4.

$\therefore P(E_7) = \dfrac{\text{No. of favourable outcomes to } E_7}{\text{Total no. of possible outcomes}} = \dfrac{4}{6} = \dfrac{2}{3}.$

Hence, the probability of getting a number divisible by 2 or 3 is $\dfrac{2}{3}$.

A die has 6 faces marked by the given numbers as shown below :

$\boxed{\phantom{1}1\phantom{1}}\space\boxed{\phantom{1}2\phantom{1}}\space\boxed{\phantom{1}3\phantom{1}}\space\boxed{-1}\space\boxed{-2}\space\boxed{-3}$

The die is thrown once. What is the probability of getting

(i) a positive integer

(ii) an integer greater than -3

(iii) the smallest integer ?

Answer

On a single throw of die,

Sample space = {1, 2, 3, -1, -2, -3}.

(i) Let E₁ be the event of getting a positive integer, then

E₁ = {1, 2, 3}.

∴ The number of favourable outcomes to the event E₁ = 3.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}.$

Hence, the probability of getting a positive integer is $\dfrac{1}{2}$.

(ii) Let E₂ be the event of getting a integer greater than -3, then

E₂ = {-2, -1, 1, 2, 3}.

∴ The number of favourable outcomes to the event E₂ = 5.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{5}{6}.$

Hence, the probability of getting a integer greater than -3 is $\dfrac{5}{6}$.

(iii) Let E₃ be the event of getting smallest integer, then

E₃ = {-3}.

∴ The number of favourable outcomes to the event E₃ = 1.

$\therefore P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{1}{6}.$

Hence, the probability of getting smallest integer is $\dfrac{1}{6}$.

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (shown in the adjoining figure) and these are equally likely outcomes. What is the probability that it will point at

(i) 8?

(ii) an odd number?

(iii) a number greater than 2?

(iv) a number less than 9?

Answer

Sample space = {1, 2, 3, 4, 5, 6, 7, 8}.

(i) Let E₁ be the event of getting 8, then

E₁ = {8}.

∴ The number of favourable outcomes to the event E₁ = 1.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{1}{8}.$

Hence, the probability of getting 8 is $\dfrac{1}{8}$.

(ii) Let E₂ be the event of getting an odd number, then

E₂ = {1, 3, 5, 7}.

∴ The number of favourable outcomes to the event E₂ = 4.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{4}{8} = \dfrac{1}{2}.$

Hence, the probability of getting an odd number is $\dfrac{1}{2}$.

(iii) Let E₃ be the event of getting a number greater than 2, then

E₃ = {3, 4, 5, 6, 7, 8}.

∴ The number of favourable outcomes to the event E₃ = 6.

$\therefore P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{6}{8} = \dfrac{3}{4}.$

Hence, the probability of getting a number greater than 2 is $\dfrac{3}{4}$.

(iv) Let E₄ be the event of getting a number less than 9, then

E₄ = {1, 2, 3, 4, 5, 6, 7, 8}.

∴ The number of favourable outcomes to the event E₄ = 8.

$\therefore P(E_4) = \dfrac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \dfrac{8}{8} = 1.$

Hence, the probability of of getting a number less than 9 is 1.

Find the probability that the month of January may have 5 Mondays in

(i) a leap year

(ii) a non-leap year

Answer

In January there are 31 days and in an ordinary year there are 365 days but in a leap year there are 366 days.

(i) In January of a leap year, there are 31 days i.e., 4 weeks and 3 days. Therefore we have to find the probability of having a Monday out of the remaining 3 days.

Now 3 days can be (Monday, Tuesday, Wednesday), (Tuesday, Wednesday, Thursday), (Wednesday, Thursday, Friday), (Thursday, Friday, Saturday), (Friday, Saturday, Sunday), (Saturday, Sunday, Monday), (Sunday, Monday, Tuesday).

In above 7 pairs, 3 times Monday occurs,

∴ Probability(having 5 Mondays) = $\dfrac{3}{7}$.

Hence, the probability that the month of January may have 5 Mondays in a leap year is $\dfrac{3}{7}$.

(ii) In January of an ordinary year, there are 31 days i.e. 4 weeks and 3 days. Out of remaining 3 days any one can be a monday,

Now 3 days can be (Monday, Tuesday, Wednesday), (Tuesday, Wednesday, Thursday), (Wednesday, Thursday, Friday), (Thursday, Friday, Saturday), (Friday, Saturday, Sunday), (Saturday, Sunday, Monday), (Sunday, Monday, Tuesday).

In above 7 pairs, 3 times Monday occurs,

∴ Probability (having 5 Mondays) = $\dfrac{3}{7}$.

Hence, the probability that the month of January may have 5 Mondays in a non-leap year is $\dfrac{3}{7}$.

Find the probability that the month of February may have 5 Wednesdays in

(i) a leap year

(ii) a non-leap year

Answer

In the month of February there are 29 days in a leap year while 28 days in a non-leap year.

(i) In a leap year's February there are 29 days i.e. 4 weeks and 1 day. In order to have 5 wednesdays, the remaining 1 day should be a wednesday

∴ Probability (having 5 wednesdays) = $\dfrac{1}{7}$.

Hence, the probability that the month of February may have 5 Wednesdays in a leap year is $\dfrac{1}{7}$.

(ii) In a non-leap year's February there are 28 days i.e. 4 weeks and 0 day.

∴ Probability (having 5 wednesdays) = $\dfrac{0}{7}$ = 0.

Hence, the probability that the month of February may have 5 Wednesdays in a non-leap year is 0.

Sixteen cards are labelled as a, b, c, ....., m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is :

(i) a vowel

(ii) a consonant

(iii) none of the letters of the word median.

Answer

On drawing a card from the box,

Sample space = {a, b, c, ......, m, n, o, p}.

(i) Let E₁ be the event of drawing a vowel card.

E₁ = {a, e, i, o}

∴ The number of favourable outcomes to the event E₁ = 4.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{4}{16} = \dfrac{1}{4}.$

Hence, the probability of drawing a vowel card is $\dfrac{1}{4}$.

(ii) Let E₂ be the event of drawing a consonant card.

Since, there are 4 vowels in the range a, b, ....., p. Hence, no. of consonants = 16 - 4 = 12.

∴ The number of favourable outcomes to the event E₂ = 12.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{12}{16} = \dfrac{3}{4}.$

Hence, the probability of drawing a consonant card is $\dfrac{3}{4}$.

(iii) Let E₃ be the event of drawing a card not containing letters of word median.

Since, there are 6 letters in the word median. Hence, no. of other letters = 16 - 6 = 10.

∴ The number of favourable outcomes to the event E₃ = 10.

$\therefore P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{10}{16}= \dfrac{5}{8}.$

Hence, the probability of drawing a card not containing letters of word 'median' is $\dfrac{5}{8}$.

Each of the letter of the word 'BOUNDARIES' is written on identical cards and put in the bag. they are shuffled if a card is drawn at random, What is the probability that the letter is

(i) a consonant

(ii) one of the letter of the word 'LUCKNOW' ?

(iii) one of the letter of the word 'INDIA' ?

Answer

(i) No. of consonants in the word 'BOUNDARIES' = 5 [B, N, D, R, S]

∴ No. of favourable outcomes = 5

Total no. of letters in the word 'BOUNDARIES' = 10.

∴ No. of possible outcomes = 10

P(that card drawn has a consonant) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{10} = \dfrac{1}{2}$

Hence, probability of getting a consonant = $\dfrac{1}{2}$.

(ii) No. of different letters in the word 'LUCKNOW' that are also present in the word 'BOUNDARIES'` = 3 (U, N, O)

∴ No. of favourable outcomes = 3

P(that the letter on card is a letter of the word 'LUCKNOW') = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{10}$

Hence, probability of getting one of the letter of the word 'LUCKNOW' = $\dfrac{3}{10}$.

(iii) No. of different letters in the word 'INDIA' that are also present in the word 'BOUNDARIES'` = 4 (I, N, D, A)

∴ No. of favourable outcomes = 4

P(that the letter on card is a letter of the word 'INDIA') = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{10} = \dfrac{2}{5}$

Hence, probability of getting one of the letter of the word 'INDIA' = $\dfrac{2}{5}$.

An integer is chosen between 0 and 100. What is the probability that it is

(i) divisible by 7?

(ii) not divisible by 7?

Answer

On choosing an integer between 0 and 100,

Sample space = {1, 2, 3, 4, ........, 99}.

(i) Let E₁ be the event of choosing a number that is divisible by 7.

E₁ = {7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}.

∴ The number of favourable outcomes to the event E₁ = 14.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{14}{99}.$

Hence, the probability of choosing an integer divisible by 7 is $\dfrac{14}{99}$.

(ii) Let E₂ be the event of choosing a number that is not divisible by 7.

Since there are 14 numbers between 0 and 100 that are divisible by 7, hence the numbers not divisible by 7 = 99 - 14 = 85.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{85}{99}.$

Hence, the probability of choosing an integer not divisible by 7 is $\dfrac{85}{99}.$

Cards marked with numbers 1, 2, 3, 4, ......, 20 are well-shuffled and a card is drawn at random. What is the probability that the number on the card is :

(i) a prime number

(ii) divisible by 3

(iii) a perfect square?

Answer

Cards are marked 1, 2, 3, 4, ......, 20 and a card is drawn at random.

Sample space = {1, 2, 3, 4, ......, 20}, which has 20 equally likely outcomes.

(i) Let E₁ be the event of choosing a prime number.

E₁ = {2, 3, 5, 7, 11, 13, 17, 19}.

∴ The number of favourable outcomes to the event E₁ = 8.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{8}{20} = \dfrac{2}{5}.$

Hence, the probability of choosing a prime number is $\dfrac{2}{5}$.

(ii) Let E₂ be the event of choosing a number that is divisible by 3.

E₂ = {3, 6, 9, 12, 15, 18}.

∴ The number of favourable outcomes to the event E₂ = 6.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{6}{20} = \dfrac{3}{10}.$

Hence, the probability of choosing a number divisible by 3 is $\dfrac{3}{10}$.

(iii) Let E₃ be the event of choosing a perfect square.

E₃ = {1, 4, 9, 16}.

∴ The number of favourable outcomes to the event E₃ = 4.

$\therefore P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{4}{20} = \dfrac{1}{5}.$

Hence, the probability of choosing a perfect square is $\dfrac{1}{5}$.

There are 25 discs numbered 1 to 25. They are put in a closed box and shaken throughly. A disc is drawn at random from the box. Find the probability that the number on the disc is:

(i) an odd number

(ii) divisible by 2 and 3 both

(iii) a number less than 16.

Answer

(i) Let E₁ be the event of choosing an odd number disc.

E₁ = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}.

∴ The number of favourable outcomes to the event E₁ = 13.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{13}{25}.$

Hence, the probability of choosing an odd number disc is $\dfrac{13}{25}$.

(ii) Let E₂ be the event of choosing a disc with number that is divisible by both 2 and 3.

E₂ = {6, 12, 18, 24}.

∴ The number of favourable outcomes to the event E₂ = 4.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{4}{25}.$

Hence, the probability of choosing a disc with number that is divisible by both 2 and 3 is $\dfrac{4}{25}$.

(iii) Let E₃ be the event of choosing a disc with number less than 16.

E₃ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}.

∴ The number of favourable outcomes to the event E₃ = 15.

$\therefore P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{15}{25} = \dfrac{3}{5}.$

Hence, the probability of choosing a disc with number less than 16 is $\dfrac{3}{5}$.

A box contains 15 cards numbered 1, 2, 3, ...., 15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is :

(i) odd

(ii) prime

(iii) divisible by 3

(iv) divisible by 3 and 2 both

(v) divisible by 3 or 2

(vi) a perfect square number.

Answer

(i) Let E₁ be the event of choosing an odd number card.

E₁ = {1, 3, 5, 7, 9, 11, 13, 15}.

∴ The number of favourable outcomes to the event E₁ = 8.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{8}{15}.$

Hence, the probability of choosing an odd number card is $\dfrac{8}{15}$.

(ii) Let E₂ be the event of choosing a prime number card.

E₂ = {2, 3, 5, 7, 11, 13}.

∴ The number of favourable outcomes to the event E₂ = 6.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{6}{15} = \dfrac{2}{5}.$

Hence, the probability of choosing a prime number card is $\dfrac{2}{5}$.

(iii) Let E₃ be the event of choosing card with number that is divisible by 3.

E₃ = {3, 6, 9, 12, 15}.

∴ The number of favourable outcomes to the event E₃ = 5.

$\therefore P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{5}{15} = \dfrac{1}{3}.$

Hence, the probability of choosing a card with number that is divisible by 3 is $\dfrac{1}{3}$.

(iv) Let E₄ be the event of choosing card with number that is divisible by 3 and 2.

E₄ = {6, 12}.

∴ The number of favourable outcomes to the event E₄ = 2.

$\therefore P(E_4) = \dfrac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \dfrac{2}{15}.$

Hence, the probability of choosing a card with number that is divisible by 3 and 2 is $\dfrac{2}{15}$.

(v) Let E₅ be the event of choosing card with number that is divisible by 3 or 2.

E₅ = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15}.

∴ The number of favourable outcomes to the event E₅ = 10.

$\therefore P(E_5) = \dfrac{\text{No. of favourable outcomes to } E_5}{\text{Total no. of possible outcomes}} = \dfrac{10}{15} = \dfrac{2}{3}.$

Hence, the probability of choosing a card with number that is divisible by 3 or 2 is $\dfrac{2}{3}$.

(vi) Let E₆ be the event of choosing card with perfect square number.

E₆ = {1, 4, 9}.

∴ The number of favourable outcomes to the event E₆ = 3.

$\therefore P(E_6) = \dfrac{\text{No. of favourable outcomes to } E_6}{\text{Total no. of possible outcomes}} = \dfrac{3}{15} = \dfrac{1}{5}.$

Hence, the probability of choosing a card with perfect square number is $\dfrac{1}{5}$.

Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card which is :

(i) a prime number

(ii) a number divisible by 4

(iii) a number that is a multiple of 6

(iv) an odd number.

Answer

Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. Hence, total no. of cards = 10.

(i) Let E₁ be the event of choosing a prime number card.

E₁ = {2}.

∴ The number of favourable outcomes to the event E₁ = 1.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{1}{10}.$

Hence, the probability of choosing a prime number card is $\dfrac{1}{10}$.

(ii) Let E₂ be the event of choosing a card with number that is divisible by 4.

E₂ = {4, 8, 12, 16, 20}.

∴ The number of favourable outcomes to the event E₂ = 5.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{5}{10} = \dfrac{1}{2}.$

Hence, the probability of choosing a card with number that is divisible by 4 is $\dfrac{1}{2}$.

(iii) Let E₃ be the event of choosing card with number that is a multiple of 6.

E₃ = {6, 12, 18}.

∴ The number of favourable outcomes to the event E₃ = 3.

$\therefore P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{3}{10}.$

Hence, the probability of choosing a card with number that is multiple of 6 is $\dfrac{3}{10}$.

(iv) Let E₄ be the event of choosing card with odd number.

E₄ = {}.

∴ The number of favourable outcomes to the event E₄ = 0.

$\therefore P(E_4) = \dfrac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \dfrac{0}{10} = 0.$

Hence, the probability of choosing a card with odd number is 0.

Cards marked with numbers 13, 14, 15, ...., 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on card drawn is

(i) divisible by 5

(ii) a perfect square number.

Answer

The cards are mixed thoroughly and a card is drawn at random from the box means that all the outcomes are equally likely.

Sample space = {13, 14, 15, ...., 60}, which has 48 equally likely outcomes.

(i) Let E₁ be the event of choosing card with number that is divisible by 5.

E₁ = {15, 20, 25, 30, 35, 40, 45, 50, 55, 60}.

∴ The number of favourable outcomes to the event E₁ = 10.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{10}{48} = \dfrac{5}{24}.$

Hence, the probability of choosing a card with number that is divisible by 5 is $\dfrac{5}{24}$.

(ii) Let E₂ be the event of choosing card with perfect square number.

E₂ = {16, 25, 36, 49}.

∴ The number of favourable outcomes to the event E₂ = 4.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{4}{48} = \dfrac{1}{12}.$

Hence, the probability of choosing a card with perfect square number is $\dfrac{1}{12}$.

Tickets numbered 3, 5, 7, 9, ...., 29 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on ticket is

(i) a prime number

(ii) a number less than 16

(iii) a number divisible by 3.

Answer

Tickets are mixed thoroughly and a ticket is drawn at random from the box means that all the outcomes are equally likely.

Sample space = {3, 5, 7, 9, ...., 29}, which has 14 equally likely outcomes.

(i) Let E₁ be the event of choosing ticket with prime number.

E₁ = {3, 5, 7, 11, 13, 17, 19, 23, 29}.

∴ The number of favourable outcomes to the event E₁ = 9.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{9}{14}.$

Hence, the probability of choosing a ticket with prime number is $\dfrac{9}{14}$.

(ii) Let E₂ be the event of choosing ticket with number less than 16.

E₂ = {3, 5, 7, 9, 11, 13, 15}.

∴ The number of favourable outcomes to the event E₂ = 7.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{7}{14} = \dfrac{1}{2}.$

Hence, the probability of choosing a ticket with number less than 16 is $\dfrac{1}{2}$.

(iii) Let E₃ be the event of choosing ticket with number divisible by 3.

E₃ = {3, 9, 15, 21, 27}.

∴ The number of favourable outcomes to the event E₃ = 5.

$\therefore P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{5}{14}.$

Hence, the probability of choosing a ticket with number divisible by 3 is $\dfrac{5}{14}$.

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5

(iv) a prime number less than 30.

Answer

A disc is drawn at random from the box means that all the outcomes are equally likely.

Sample space = {1, 2, 3, ....., 90}, which has 90 equally likely outcomes.

(i) Let E₁ be the event of drawing a disc with two digit number.

E₁ = {10, 11, 12, 13, ......, 90}.

∴ The number of favourable outcomes to the event E₁ = 81.

$\therefore P(E_1) = \dfrac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \dfrac{81}{90} = \dfrac{9}{10}.$

Hence, the probability of drawing a disc with two digit number is $\dfrac{9}{10}.$

(ii) Let E₂ be the event of drawing a disc with perfect square number.

E₂ = {1, 4, 9, 16, 25, 36, 49, 64, 81}.

∴ The number of favourable outcomes to the event E₂ = 9.

$\therefore P(E_2) = \dfrac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \dfrac{9}{90} = \dfrac{1}{10}.$

Hence, the probability of drawing a disc with perfect square number is $\dfrac{1}{10}$.

(iii) Let E₃ be the event of drawing a disc with number divisible by 5.

E₃ = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90}.

∴ The number of favourable outcomes to the event E₃ = 18.

$\therefore P(E_3) = \dfrac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \dfrac{18}{90} = \dfrac{1}{5}.$

Hence, the probability of drawing a disc with number that is divisible by 5 is $\dfrac{1}{5}$.

(iv) Let E₄ be the event of drawing a disc with prime number less than 30.

E₄ = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}.

∴ The number of favourable outcomes to the event E₄ = 10.

$\therefore P(E_4) = \dfrac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \dfrac{10}{90} = \dfrac{1}{9}.$

Hence, the probability of drawing a disc with prime number less than 30 is $\dfrac{1}{9}$.

A bag contains 15 balls of which some are white and others are red. If the probability of drawing a red ball is twice that of a white ball, find the number of white balls in the bag.

Answer

Let no. of white balls in bag = x, so no. of red balls = 15 - x.

P(drawing a white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{Total outcomes}} = \dfrac{x}{15}$

P(drawing a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{Total outcomes}} = \dfrac{15 - x}{15}$.

Given, P(drawing a red ball) = 2 × P(drawing a white ball)

$\therefore \dfrac{15 - x}{15} = \dfrac{2x}{15} \\[1em] \Rightarrow 15 - x = 2x \\[1em] \Rightarrow 2x + x = 15 \\[1em] \Rightarrow 3x = 15 \\[1em] \Rightarrow x = 5.$

Hence, there are 5 white balls in the bag.

A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball is twice that of a red ball, find the number of balls in the bag.

Answer

Let no. of blue balls be x, total no. of balls = x + 6.

P(drawing a blue ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{Total outcomes}} = \dfrac{x}{x + 6}$

P(drawing a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{Total outcomes}} = \dfrac{6}{x + 6}$.

Given, P(drawing a blue ball) = 2 × P(drawing a red ball)

$\therefore \dfrac{x}{x + 6} = 2 \times \dfrac{6}{x + 6} \\[1em] \Rightarrow \dfrac{x}{x + 6} = \dfrac{12}{x + 6} \\[1em] \Rightarrow x = 12.$

Total no. of balls = x + 6 = 12 + 6 = 18.

Hence, there are 18 balls in the bag.

A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. Find the probability that it is

(i) white

(ii) not red.

Answer

Total balls = x + 2x + 3x = 24.

⇒ 6x = 24
⇒ x = 4.

(i) P(drawing a white ball) = $\dfrac{\text{No. of white balls}}{\text{Total balls}} = \dfrac{2 \times 4}{24} = \dfrac{1}{3}.$

Hence, the probability of drawing a white ball is $\dfrac{1}{3}$.

(ii) Total no. of white and blue balls = 2x + 3x = 5x = 20.

P(A) = P(drawing a not red ball) = P(drawing a white or blue ball).

$\therefore P(A) = \dfrac{\text{Total no. of white and blue balls}}{\text{Total balls}} \\[1em] = \dfrac{20}{24} \\[1em] = \dfrac{5}{6}.$

Hence, the probability of not drawing a red ball is $\dfrac{5}{6}$.

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting :

(i) '2' of spades

(ii) a jack

(iii) a king of red colour

(iv) a card of diamond

(v) a king or a queen

(vi) a non-face card

(vii) a black face card

(viii) a black card

(ix) a non-ace

(x) non-face card of black colour

(xi) neither a spade nor a jack

(xii) neither a heart nor a red king.

Answer

Well-shuffling ensures equally likely outcomes.

Total number of outcomes = 52.

(i) There is only one 2 of spades in the whole pack.

∴ P('2' of spades) = $\dfrac{1}{52}$.

Hence, the probability of drawing 2 of spades = $\dfrac{1}{52}$.

(ii) There are 4 jacks, one of each suit.

∴ The number of favourable outcomes to the event 'a jack' = 4.

∴ P(a jack) = $\dfrac{4}{52} = \dfrac{1}{13}$.

Hence, the probability of drawing a jack = $\dfrac{1}{13}$.

(iii) There are two king of red colour, one of hearts and one of diamond.

∴ The number of favourable outcomes to the event 'king of red colour' = 2.

∴ P(king of red colour) = $\dfrac{2}{52} = \dfrac{1}{26}.$

Hence, the probability of drawing a king of red colour = $\dfrac{1}{26}$.

(iv) There are 13 cards of diamond suit.

∴ The number of favourable outcomes to the event 'a card of diamond' = 13.

∴ P(a card of diamond) = $\dfrac{13}{52} = \dfrac{1}{4}$.

Hence, the probability of drawing a diamond card = $\dfrac{1}{4}$.

(v) There are 8 king and queen cards, 2 in each suit.

∴ The number of favourable outcomes to the event 'a king or queen' = 8.

∴ P(a king or queen) = $\dfrac{8}{52} = \dfrac{2}{13}$.

Hence, the probability of drawing a king or queen = $\dfrac{2}{13}$.

(vi) There are 12 face cards.

∴ No. of non-face cards = 52 - 12 = 40.

∴ The number of favourable outcomes to the event 'a non-face card' = 40.

∴ P(a non-face card) = $\dfrac{40}{52} = \dfrac{10}{13}$.

Hence, the probability of drawing a non-face card = $\dfrac{10}{13}$.

(vii) Since 2 suits are of black colour and each suit has 3 face cards.

∴ No. of black face cards = 2 × 3 = 6.

∴ The number of favourable outcomes to the event 'a black face card' = 6.

∴ P(a black face card) = $\dfrac{6}{52} = \dfrac{3}{26}$.

Hence, the probability of drawing a black face card = $\dfrac{3}{26}$.

(viii) There are 2 suits of black cards.

∴ No. of black cards = 26.

∴ The number of favourable outcomes to the event 'a black card' = 26.

∴ P(a black card) = $\dfrac{26}{52} = \dfrac{1}{2}$.

Hence, the probability of drawing a black card = $\dfrac{1}{2}$.

(ix) There are 4 ace cards, one in each suit.

∴ No. of non-ace cards = 52 - 4 = 48.

∴ The number of favourable outcomes to the event 'a non-ace card' = 48.

∴ P(a non-ace card) = $\dfrac{48}{52} = \dfrac{12}{13}$.

Hence, the probability of drawing a non-ace card = $\dfrac{12}{13}$.

(x) There are 3 face cards in each suit and 2 suits of black colour.

Hence, no. of face cards of black colour = 6.

∴ No. of non-ace black cards = 26 - 6 = 20.

∴ The number of favourable outcomes to the event 'a non-face card of black colour' = 20.

∴ P(a non-face black card) = $\dfrac{20}{52} = \dfrac{5}{13}$.

Hence, the probability of drawing a non-face black card = $\dfrac{5}{13}$.

(xi) There are 13 spade cards and each suit has 1 jack.

So, the other 3 suits apart from spade has 3 jacks.

∴ Total no. of spade and jack cards = 13 + 3 = 16.

Hence, no. of cards other than spade and jack = 52 - 16 = 36.

∴ The number of favourable outcomes to the event 'neither a spade nor a jack' = 36.

∴ P(neither a spade nor a jack) = $\dfrac{36}{52} = \dfrac{9}{13}$.

Hence, the probability of drawing neither a spade nor a jack = $\dfrac{9}{13}$.

(xii) There are 13 heart cards and 2 suits of red colour.

Since, one king of red colour is already included in hearts hence only one red king is more.

∴ Total no. of heart and red king cards = 13 + 1 = 14.

Hence, no. of cards other than heart and red king cards = 52 - 14 = 38.

∴ The number of favourable outcomes to the event 'neither heart nor a red king' = 38.

∴ P(neither a heart nor a red king) = $\dfrac{38}{52} = \dfrac{19}{26}$.

Hence, the probability of drawing neither a heart nor a red king = $\dfrac{19}{26}$.

All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting

(i) a black face card

(ii) a queen

(iii) a black card

(iv) a heart

(v) a spade

(vi) '9' of black colour.

Answer

3 face cards of spades are removed, hence total cards = 52 - 3 = 49.

Well-shuffling ensures equally likely outcomes.

Total number of outcomes = 49.

(i) Since 2 suits are of black colour and each suit has 3 face cards but since spades of black colour are removed.

∴ No. of black face cards = 3.

∴ The number of favourable outcomes to the event 'a black face card' = 3.

∴ P(a black face card) = $\dfrac{3}{49}$.

Hence, the probability of drawing a black face card = $\dfrac{3}{49}$.

(ii) Each suit has one queen.

Since, face cards of spades are removed hence, it has no queen.

So, there are 3 queens left.

∴ The number of favourable outcomes to the event 'a queen' = 3.

∴ P(a queen) = $\dfrac{3}{49}$.

Hence, the probability of drawing a queen = $\dfrac{3}{49}$.

(iii) Total no. of black cards = 26.

Since, spades are of black colour and it's face card are removed.

∴ The number of black cards left = 26 - 3 = 23.

∴ The number of favourable outcomes to the event 'a black card' = 23.

∴ P(a black card) = $\dfrac{23}{49}$.

Hence, the probability of drawing a black card = $\dfrac{23}{49}$.

(iv) There are 13 heart cards.

∴ The number of favourable outcomes to the event 'a heart' = 13.

∴ P(a heart) = $\dfrac{13}{49}$.

Hence, the probability of drawing a heart = $\dfrac{13}{49}$.

(v) Since, face cards of spades are removed,

∴ No. of spades left = 13 - 3 = 10.

∴ P(a spade) = $\dfrac{10}{49}$.

Hence, the probability of drawing a spade = $\dfrac{10}{49}$.

(vi) There are 2, '9' numbered cards of black colour.

∴ P(a '9' of black colour) = $\dfrac{2}{49}$.

Hence, the probability of drawing a '9' of black colour = $\dfrac{2}{49}$.

From a pack of 52 cards, a black jack, a red queen and two black kings fell down. A card was then drawn from the remaining pack at random. Find the probability that the card drawn is

(i) a black card

(ii) a king

(iii) a red queen.

Answer

Given, a black jack, a red queen and two black kings fell down.

Hence, remaining no. of cards = 52 - 4 = 48.

(i) There are total 26 black cards, 13 of club and 13 of spades. Out of which 3 are removed.

∴ No. of black cards left = 26 - 3 = 23.

P(a black card) = $\dfrac{23}{48}.$

Hence, the probability of drawing a black card is $\dfrac{23}{48}.$

(ii) There are total 4 kings, but 2 black kings are removed.

∴ No. of kings left = 4 - 2 = 2.

P(a king) = $\dfrac{2}{48} = \dfrac{1}{24}.$

Hence, the probability of drawing a king is $\dfrac{1}{24}.$

(iii) There are 2 red queens, one of hearts and one of diamonds.

Since, one red queen is removed, hence no. of red queens left = 2 - 1 = 1.

P(a red queen) = $\dfrac{1}{48}.$

Hence, the probability of drawing a red queen is $\dfrac{1}{48}.$

Two coins are tossed once. Find the probability of getting :

(i) 2 heads

(ii) atleast one tail.

Answer

When two different coins are tossed simultaneously, then sample space = {HH, HT, TH, TT}. It consist of 4 equally likely outcomes.

Total number of possible outcomes = 4.

(i) Let A be the event 'two heads', then A = {HH} and the number of outcomes favourable to the event A = 1.

∴ P(two heads) = $\dfrac{1}{4}$.

Hence, the probability of getting two heads = $\dfrac{1}{4}$.

(ii) Let B be the event 'atleast one tail', then B = {HT, TT, TH} and the number of outcomes favourable to the event B = 3.

∴ P(atleast one tail) = $\dfrac{3}{4}$.

Hence, the probability of getting atleast one tail = $\dfrac{3}{4}$.

Two different coins are tossed simultaneously. Find the probability of getting :

(i) two tails

(ii) one tail

(iii) no tail

(iv) atmost one tail.

Answer

When two different coins are tossed simultaneously, then sample space = {HH, HT, TH, TT}. It consists of 4 equally likely outcomes.

Total number of possible outcomes = 4.

(i) Let A be the event 'two tails', then A = {TT} and the number of outcomes favourable to the event A = 1.

∴ P(two tails) = $\dfrac{1}{4}$.

Hence, the probability of getting two tails = $\dfrac{1}{4}$.

(ii) Let B be the event 'one tail', then A = {HT, TH} and the number of outcomes favourable to the event B = 2.

∴ P(one tail) = $\dfrac{2}{4} = \dfrac{1}{2}$.

Hence, the probability of getting one tail = $\dfrac{1}{2}$.

(iii) Let C be the event 'no tail', then C = {HH} and the number of outcomes favourable to the event C = 1.

∴ P(no tail) = $\dfrac{1}{4}$.

Hence, the probability of getting no tail = $\dfrac{1}{4}$.

(iv) Let D be the event 'atmost one tail', then D = {HH, HT, TH} and the number of outcomes favourable to the event D = 3.

∴ P(atmost one tail) = $\dfrac{3}{4}$.

Hence, the probability of getting atmost one tail = $\dfrac{3}{4}$.

Two different dice are thrown simultaneously. Find the probability of getting :

(i) a number greater than 3 on each dice

(ii) an odd number on both dice.

Answer

When two different dice are rolled together, the total number of outcomes is 6 × 6 i.e. 36 and all outcomes are equally likely. The sample space of the random experiment has 36 equally likely outcomes. The sample of the experiment

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
       (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
       (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
       (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
       (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
       (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).}

It consists of 36 equally likely outcomes.

(i) Let A be the event of getting 'a number greater than 3 on each dice', then

A = {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}.

∴ The number of outcomes favourable to event A = 9.

∴ P(a number greater than 3 on each dice) = $\dfrac{9}{36} = \dfrac{1}{4}.$

Hence, the probability of getting a number greater than 3 on each dice is $\dfrac{1}{4}$.

(ii) Let B be the event of getting 'an odd number on both dice', then

A = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}.

∴ The number of outcomes favourable to event B = 9.

∴ P(an odd number on both the dice) = $\dfrac{9}{36} = \dfrac{1}{4}.$

Hence, the probability of getting an odd number on both the dice is $\dfrac{1}{4}$.

Two different dice are thrown at the same time. Find the probability of getting :

(i) a doublet

(ii) a sum of 8

(iii) sum divisible by 5

(iv) sum of atleast 11.

Answer

(i) Let A be the event of getting 'a doublet', then

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}.

∴ The number of outcomes favourable to event A = 6.

∴ P(a doublet) = $\dfrac{6}{36} = \dfrac{1}{6}.$

Hence, the probability of getting a doublet is $\dfrac{1}{6}$.

(ii) Let B be the event of getting 'a sum of 8', then

A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}.

∴ The number of outcomes favourable to event B = 5.

∴ P(a sum of 8) = $\dfrac{5}{36}.$

Hence, the probability of getting a sum of 8 is $\dfrac{5}{36}$.

(iii) Let C be the event of getting 'a sum divisible by 5', then

C = {(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)}.

∴ The number of outcomes favourable to event C = 7.

∴ P(a sum divisible by 5) = $\dfrac{7}{36}.$

Hence, the probability of getting a sum divisible by 5 is $\dfrac{7}{36}$.

(iv) Let D be the event of getting 'sum of atleast 11', then

D = {(5, 6), (6, 5), (6, 6)}.

∴ The number of outcomes favourable to event D = 3.

∴ P(sum of atleast 11) = $\dfrac{3}{36} = \dfrac{1}{12}.$

Hence, the probability of getting a sum of atleast 11 is $\dfrac{1}{12}$.

The following letters A, D, M, N, O, S, U, Y of the English alphabet are written on separate cards and put in a box. The cards are well shuffled and one card is drawn at random. What is the probability that the card drawn is a letter of the word,

(a) MONDAY?

(b) which does not appear in MONDAY?

(c) which appears both in SUNDAY and MONDAY?

Answer

Letters written on cards = {'A', 'D', 'M', 'N', 'O', 'S', 'U', 'Y'}

No. of cards = 8

(a) Letters of the word MONDAY present in the cards = {'M', 'O', 'N', 'D', 'A', 'Y'}

Probability that the card drawn is a letter of the word MONDAY

= $\dfrac{\text{Letters of MONDAY present}}{\text{No. of cards}} = \dfrac{6}{8} = \dfrac{3}{4}$.

Hence, required probability = $\dfrac{3}{4}$.

(b) Letters of the word not present in MONDAY = {'S', 'U'}

Probability that the card drawn is not a letter of the word MONDAY

= $\dfrac{\text{Letters not present in MONDAY}}{\text{No. of cards}} = \dfrac{2}{8} = \dfrac{1}{4}$.

Hence, required probability = $\dfrac{1}{4}$.

(c) Letters of the word present in SUNDAY and MONDAY are {'N', 'D', 'A', 'Y'}.

Probability that the card drawn has a letter which appears both in SUNDAY and MONDAY

= $\dfrac{\text{Letters present in both MONDAY and SUNDAY}}{\text{No. of cards}} = \dfrac{4}{8} = \dfrac{1}{2}$.

Hence, required probability = $\dfrac{1}{2}$.

In a T.V. show, a contestant opt for video call a friend life line to get an answer from three of his friends, named Amar, Akbar and Anthony. The question which he asks from one of his friends has four options. Find the probability that :

(a) Akbar is chosen for the call.

(b) Akbar couldn't give the correct answer.

Answer

(a) Since, there are three people which can be called.

Hence, probability that Akbar is chosen for the call = $\dfrac{1}{3}$.

(b) Since, there are four options out of which one is correct.

Hence, probability that Akbar couldn't give the correct answer = $\dfrac{3}{4}$.

Which of the following cannot be the probability of an event?

1. 0.7

2. $\dfrac{2}{3}$

3. -1.5

4. 15%

Answer

Since, the probability cannot be negative.

Hence, Option 3 is the correct option.

If the probability of an event is p, then the probability of its complementary event will be

1. p - 1

2. p

3. 1 - p

4. 1 - $\dfrac{1}{p}$

Answer

Probability of an event is p.

∴ Probability of complementary event is 1 - p.

Hence, Option 3 is the correct option.

Out of one digit prime numbers, one number is selected at random. The probability of selecting an even number is

1. $\dfrac{1}{2}$

2. $\dfrac{1}{4}$

3. $\dfrac{4}{9}$

4. $\dfrac{2}{5}$

Answer

Sample space = {2, 3, 5, 7}.

Let A be the event of selecting an even number,

∴ A = {2}.

Hence, the no. of favourable outcomes to A = 1.

∴ P(A) = $\dfrac{1}{4}.$

Hence, Option 2 is the correct option.

When a die is thrown, the probability of getting an odd number less than 3 is

1. $\dfrac{1}{6}$

2. $\dfrac{1}{3}$

3. $\dfrac{1}{2}$

4. 0

Answer

When a die is thrown the sample space is,

S = {1, 2, 3, 4, 5, 6}.

Let A be the event of getting an odd number less than 3,

∴ A = {1}.

Hence, the no. of favourable outcomes to A = 1.

∴ P(A) = $\dfrac{1}{6}.$

Hence, Option 1 is the correct option.

The probability of getting a number divisible by 3 in throwing a die is

1. $\dfrac{1}{6}$

2. $\dfrac{1}{3}$

3. $\dfrac{1}{2}$

4. $\dfrac{2}{3}$

Answer

There are 2 numbers between 1 to 6 that are divisible by 3, i.e., 3 and 6.

∴ No. of favourable outcomes = 2

Total no. in a dice = 6.

∴ No. of possible outcomes = 6

P(a number divisible by 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}$.

Hence, option 2 is the correct option.

A fair die is thrown once. The probability of getting an even prime number is

1. $\dfrac{1}{6}$

2. $\dfrac{2}{3}$

3. $\dfrac{1}{3}$

4. $\dfrac{1}{2}$

Answer

When a die is thrown the sample space is,

S = {1, 2, 3, 4, 5, 6}.

Let A be the event of getting an even prime number,

∴ A = {2}.

Hence, the no. of favourable outcomes to A = 1.

∴ P(A) = $\dfrac{1}{6}.$

Hence, Option 1 is the correct option.

A fair die is thrown once. The probability of getting a composite number is

1. $\dfrac{1}{3}$

2. $\dfrac{1}{6}$

3. $\dfrac{2}{3}$

4. 0

Answer

When a die is thrown the sample space is,

S = {1, 2, 3, 4, 5, 6}.

Let A be the event of getting a composite number,

∴ A = {4, 6}.

Hence, the no. of favourable outcomes to A = 2.

∴ P(A) = $\dfrac{2}{6} = \dfrac{1}{3}.$

Hence, Option 1 is the correct option.

If a fair die is rolled once, then the probability of getting an even number or a number greater than 4 is

1. $\dfrac{1}{2}$

2. $\dfrac{1}{3}$

3. $\dfrac{5}{6}$

4. $\dfrac{2}{3}$

Answer

When a die is thrown the sample space is,

S = {1, 2, 3, 4, 5, 6}.

Let A be the event of getting an even number or a number greater than 4,

∴ A = {2, 4, 5, 6}.

Hence, the no. of favourable outcomes to A = 4.

∴ P(A) = $\dfrac{4}{6} = \dfrac{2}{3}.$

Hence, Option 4 is the correct option.

If a letter is chosen at random from the letters of English alphabet, then the probability that it is a letter of the word 'DELHI' is

1. $\dfrac{1}{5}$

2. $\dfrac{1}{26}$

3. $\dfrac{5}{26}$

4. $\dfrac{21}{26}$

Answer

Total no. of letters in English alphabet = 26.

Let A be the event of getting a letter of the word 'DELHI',

∴ A = {D, E, L, H, I}.

Hence, the no. of favourable outcomes to A = 5.

∴ P(A) = $\dfrac{5}{26}.$

Hence, Option 3 is the correct option.

A card is selected at random from a pack of 52 cards. The probability of its being a red face card is

1. $\dfrac{3}{26}$

2. $\dfrac{3}{13}$

3. $\dfrac{2}{13}$

4. $\dfrac{1}{2}$

Answer

There are 6 red face cards, 3 of hearts and 3 of diamonds.

∴ The number of favourable outcomes to event 'red face card' = 6.

∴ P(red face card) = $\dfrac{6}{52} = \dfrac{3}{26}$.

Hence, Option 1 is the correct option.

If a card is drawn from a well-shuffled pack of 52 playing cards, then the probability of this card being a king or jack is

1. $\dfrac{1}{26}$

2. $\dfrac{1}{13}$

3. $\dfrac{2}{13}$

4. $\dfrac{4}{13}$

Answer

Well-shuffling ensures equally likely outcomes.

Total number of outcomes = 52.

There are 4 kings and 4 jacks i.e. total 8 kings and jacks.

∴ The number of favourable outcomes to event 'being a king or jack' is = 8.

∴ P(a king or jack) = $\dfrac{8}{52} = \dfrac{2}{13}.$

Hence, Option 3 is the correct option.

The probability that a non-leap year selected at random has 53 Sundays is

1. $\dfrac{1}{365}$

2. $\dfrac{2}{365}$

3. $\dfrac{2}{7}$

4. $\dfrac{1}{7}$

Answer

Number of days in a non-leap year = 365 i.e. 52 weeks and 1 day.

In order to have 53 sundays, the 1 day should be a sunday.

Let A be the event of the day to be a Sunday,

∴ P(A) = $\dfrac{1}{7}$.

Hence, Option 4 is the correct option.

A bag contains 3 red balls, 5 white balls and 7 black balls. The probability that a ball drawn from the bag at random will be neither red nor black is

1. $\dfrac{1}{5}$

2. $\dfrac{1}{3}$

3. $\dfrac{7}{15}$

4. $\dfrac{8}{15}$

Answer

Probability that a ball drawn from the bag at random will be neither red nor black is equal to probability of picking a white ball.

No. of white balls = 5, total balls = 3 + 5 + 7 = 15.

∴ P(neither red nor black ball) = $\dfrac{5}{15} = \dfrac{1}{3}$.

Hence, Option 2 is the correct option.

A bag contains 4 red balls and 5 green balls. One ball is drawn at random from the bag. The probability of getting either a red ball or a green ball is

1. $\dfrac{4}{9}$

2. $\dfrac{5}{9}$

3. 0

4. 1

Answer

Total no. of balls = 4 + 5 = 9.

Let A be the event of drawing a red or green ball. As, the total no. of red and green balls = 9, so the number of favourable outcomes to event A = 9.

∴ P(A) = $\dfrac{9}{9}$ = 1.

Hence, Option 4 is the correct option.

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is

1. $\dfrac{1}{5}$

2. $\dfrac{3}{5}$

3. $\dfrac{4}{5}$

4. $\dfrac{1}{3}$

Answer

Sample space = {1, 2, 3, ......., 40}, with 40 equally likely outcomes.

Let A be the event that selected ticket has a number which is a multiple of 5.

∴ A = {5, 10, 15, 20, 25, 30, 35, 40}.

Hence, the number of favourable outcomes to event A = 8.

∴ P(A) = $\dfrac{8}{40} = \dfrac{1}{5}$.

Hence, Option 1 is the correct option.

If a number is randomly chosen from the numbers 1, 2, 3, 4, ........, 25, then probability of the number to be prime is

1. $\dfrac{7}{25}$

2. $\dfrac{9}{25}$

3. $\dfrac{11}{25}$

4. $\dfrac{13}{25}$

Answer

Sample space = {1, 2, 3, ......., 25}, with 25 equally likely outcomes.

Let A be the event that chosen number is prime.

∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23}.

Hence, the number of favourable outcomes to event A = 9.

∴ P(A) = $\dfrac{9}{25}$.

Hence, Option 2 is the correct option.

A box contains 90 cards numbered 1 to 90. If one card is drawn from the box at random, then the probability that the number on the card is a perfect square is

1. $\dfrac{1}{10}$

2. $\dfrac{9}{100}$

3. $\dfrac{1}{9}$

4. $\dfrac{3}{100}$

Answer

Sample space = {1, 2, 3, ...., 90}, which has 90 equally likely outcomes.

Let A be the event of drawing a perfect square card.

∴ A = {1, 4, 9, 16, 25, 36, 49, 64, 81}.

Hence, the number of favourable outcomes to event A = 9.

∴ P(A) = $\dfrac{9}{90} = \dfrac{1}{10}$.

Hence, Option 1 is the correct option.

If a (fair) coin is tossed twice, then the probability of getting two heads is

1. $\dfrac{1}{4}$

2. $\dfrac{1}{2}$

3. $\dfrac{3}{4}$

4. 0

Answer

A coin is tossed twice,

Sample space = {HH, HT, TH, TT}.

Let A be the event of getting two heads,

∴ A = {HH}.

∴ P(A) = $\dfrac{1}{4}$.

Hence, Option 1 is the correct option.

If two coins are tossed simultaneously, then the probability of getting atleast one head is

1. $\dfrac{1}{4}$

2. $\dfrac{1}{2}$

3. $\dfrac{3}{4}$

4. 1

Answer

When two different coins are tossed simultaneously, then sample space = {HH, HT, TH, TT}. It consist of 4 equally likely outcomes.

Total number of possible outcomes = 4.

Let A be the event of getting atleast one head,

∴ A = {HH, HT, TH}.

Hence, the no. of favourable outcomes to event A = 3.

∴ P(A) = $\dfrac{3}{4}$.

Hence, Option 3 is the correct option.

Lakshmi tosses two coins simultaneously. The probability that she gets atmost one head is

1. 1

2. $\dfrac{3}{4}$

3. $\dfrac{1}{2}$

4. $\dfrac{1}{7}$

Answer

When two different coins are tossed simultaneously, then sample space = {HH, HT, TH, TT}. It consist of 4 equally likely outcomes.

Total number of possible outcomes = 4.

Let A be the event of getting atmost one head,

∴ A = {TT, HT, TH}.

Hence, the no. of favourable outcomes to event A = 3.

∴ P(A) = $\dfrac{3}{4}$.

Hence, Option 2 is the correct option.

The probability of getting a bad egg in a lot of 400 eggs is 0.035. The number of bad eggs in the lot is

1. 7

2. 14

3. 21

4. 28

Answer

Let the no. of bad eggs be x.

$\therefore \text{ P(getting a bad egg)} = \dfrac{x}{400} \\[1em] \Rightarrow 0.035 = \dfrac{x}{400} \\[1em] \Rightarrow x = 0.035 \times 400 \\[1em] \Rightarrow x = 14.$

Hence, Option 2 is the correct option.

Event A : The sun will rise from east tomorrow.

Event B : It will rain on Monday.

Event C : February month has 29 days in a leap year.

Which of the above event(s) has probability equal to 1 ?

1. all events A, B and C

2. both events A and B

3. both events B and C

4. both events A and C

Answer

The sun always rises from east and also in a leap year February has 29 days.

∴ Event A and Event C has probability equal to 1.

Hence, Option 4 is the correct option.

A bag contains 3 red and 2 blue marbles. A marble is drawn at random. The probability of drawing a black marble is :

1. 0

2. $\dfrac{1}{5}$

3. $\dfrac{2}{5}$

4. $\dfrac{3}{5}$

Answer

Since, there is no black marble in the bag.

∴ Probability of drawing a black marble = 0.

Hence, Option 1 is the correct option.

Assertion (A): The probability that a leap year has 53 Sunday is $\dfrac{2}{7}$.

Reason (R): The probability that a non-leap year has 53 Sunday is $\dfrac{5}{7}$.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

In a leap year, there are 366 days.

366 days = 52 weeks + 2 days

These 2 days can be (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), and (Sun, Mon).

Total number of possible outcomes = 7

Number of favourable outcomes (Getting Sunday as one of the extra days) = 2 (i.e., (Sat, Sun), (Sun, Mon)).

P(Getting Sunday as one of the extra days) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{7}$

∴ Assertion (A) is true.

In a non - leap year, there are 365 days.

365 days = 52 weeks + 1 days

These 1 days can be Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.

Total number of possible outcomes = 7

Number of favourable outcomes (Getting Sunday as one of the extra days) = 1

P(Getting Sunday as one of the extra days) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{7}$

∴ Reason (R) is false.

∴ Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

Assertion (A): Two players Sania and Asma play a tennis match. If the probability of Sania winning the match is 0.79, then the probability of Asma winning the match is 0.21.

Reason (R): The sum of probabilities of two complementary event is 1.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given, the probability of Sania winning the match = 0.79.

As we know that the sum of probabilities of two complementary event is 1.

⇒ P(Sania wins) + P(Asma wins) = 1

∴ Reason (R) is true.

⇒ 0.79 + P(Asma wins) = 1

⇒ P(Asma wins) = 1 - P(Sania wins)

⇒ P(Asma wins) = 1 - 0.79

⇒ P(Asma wins) = 0.21

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Assertion (A): If the probability of occurrence of an event E is $\dfrac{5}{11}$, then the probability of non-occurrence of the event E is $\dfrac{7}{11}$.

Reason (R): If E is an event, then P(E) + P($\overline{E}$) = 1.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given, the probability of occurrence of an event E = $\dfrac{5}{11}$

As we know that the sum of probabilities of two complementary event is 1.

⇒ P(E) + P($\overline{E}$) = 1

∴ Reason (R) is true.

$\Rightarrow \dfrac{5}{11} + P(\overline{E}) = 1\\[1em] \Rightarrow P(\overline{E}) = 1 - \dfrac{5}{11}\\[1em] \Rightarrow P(\overline{E}) = \dfrac{11 - 5}{11}\\[1em] \Rightarrow P(\overline{E}) = \dfrac{6}{11}$

∴ Assertion (A) is false.

∴ Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Assertion (A): In a throw of two fare points once, the probability of getting one head is $\dfrac{1}{2}$.

Reason (R): In a throw of two fare coin, the sample space is HH, HT, TH, TT.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

When two coins are tossed together, the total number of possible outcomes = 4 (i.e. HH, HT, TH and TT)

∴ Reason (R) is true.

Number of favourable outcomes (Getting one head) = 2 (HT and TH)

P(getting one head) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{4} = \dfrac{1}{2}$

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Assertion (A): The probability of getting a prime number, when a die is thrown once, is $\dfrac{2}{3}$.

Reason (R): On the faces of a die, prime numbers are 2, 3, 5.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

On the faces of a die, numbers are 1, 2, 3, 4, 5 and 6. Out of which 2, 3 and 5 are only prime numbers.

∴ Reason (R) is true.

Number of favourable outcomes (of getting prime number) = 3

Number of possible outcomes = 6

P(getting a prime number) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$

∴ Assertion (A) is false.

∴ Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Assertion (A): A die is thrown once and the probability of getting an even number is $\dfrac{2}{3}$.

Reason (R): The sample space for even number on a die is {2, 4, 6}.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

On the faces of a die, numbers are 1, 2, 3, 4, 5 and 6. Out of which 2, 4 and 6 are even numbers.

So, the sample space for even number on a die = {2, 4, 6}

∴ Reason (R) is true.

Number of favourable outcomes = 3 (for getting even numbers)

The total number of possible outcomes = 6

P(getting a even number) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$.

∴ Assertion (A) is false.

∴ Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

In a single throw of die, find the probability of getting

(i) a number greater than 5

(ii) an odd prime number

(iii) a number which is multiple of 3 or 4.

Answer

When a die is thrown once, the possible outcomes are the numbers 1, 2, 3, 4, 5, 6. So, the sample space of the experiment = {1, 2, 3, 4, 5, 6}. It has six equally likely outcomes.

(i) The event is getting number greater than 5 i.e. {6}

The number of favourable outcomes to the event getting number greater than 5 = 1.

∴ P(getting number greater than 5) = $\dfrac{1}{6}$.

Hence, the probability of getting a number greater than 5 is $\dfrac{1}{6}$.

(ii) The event is getting an odd prime number i.e. {3, 5}

The number of favourable outcomes to the event an odd prime number = 2.

∴ P(getting an odd prime number) = $\dfrac{2}{6} = \dfrac{1}{3}$.

Hence, the probability of getting an odd prime number is $\dfrac{1}{3}$.

(iii) The event is getting a number which is a multiple of 3 or 4 i.e. {3, 4, 6}

The number of favourable outcomes to the above event = 3.

∴ P(getting a number which is a multiple of 3 or 4) = $\dfrac{3}{6} = \dfrac{1}{2}$.

Hence, the probability of getting a number which is a multiple of 3 or 4 is $\dfrac{1}{2}$.

A lot consist of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is

(i) acceptable to Varnika?

(ii) acceptable to the trader?

Answer

Total phones = 48.

(i) Varnika accepts good phones only, hence no. of favourable outcomes acceptable to Varnika = 42.

P(acceptable to Varnika) = $\dfrac{42}{48} = \dfrac{7}{8}.$

Hence, the probability that a phone is acceptable to Varnika is $\dfrac{7}{8}.$

(ii) Trader accepts good phones and one with minor defects only, hence no. of favourable outcomes acceptable to trader = 42 + 3 = 45.

P(acceptable to trader) = $\dfrac{45}{48} = \dfrac{15}{16}.$

Hence, the probability that a phone is acceptable to trader is $\dfrac{15}{16}.$

A bag contains 5 red, 8 white and 7 black balls. A ball is drawn from the bag at random. Find the probability that the drawn ball is

(i) red or white

(ii) not black

(iii) neither white nor black.

Answer

Since, a ball is drawn at random from the bag, so all the balls are equally likely to be drawn.

Total number of balls in the bag = 5 + 8 + 7 = 20.

So, the sample space of the experiment has 20 equally likely outcomes.

(i) Let E₁ be the event 'a red or white ball is drawn'.

The number of red or white balls = 5 + 8 = 13.

So, the number of favourable outcomes to the E₁ = 13.

∴ P(E₁) = P(a red or white ball) = $\dfrac{13}{20}.$

Hence, the probability that the ball drawn is red or white is $\dfrac{13}{20}.$

(ii) Drawing a not black ball means drawing a red or white ball.

∴ P(not black ball) = P(a red or white ball) = $\dfrac{13}{20}.$

Hence, the probability that the ball drawn is not a black ball is $\dfrac{13}{20}.$

(iii) If a ball drawn is neither white nor black it means a red ball is drawn.

No. of red balls = 5.

∴ P(a red ball is drawn) = $\dfrac{5}{20} = \dfrac{1}{4}.$

Hence, the probability that a red ball is drawn is $\dfrac{1}{4}.$

A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is :

(i) white or blue

(ii) red or black

(iii) not white

(iv) neither white nor black?

Answer

Since, a ball is drawn at random from the bag, so all the balls are equally likely to be drawn.

Total number of balls in the bag = 5 + 7 + 4 + 2 = 18.

So, the sample space of the experiment has 18 equally likely outcomes.

(i) Let E₁ be the event 'a white or blue ball is drawn'.

The number of white or blue balls = 5 + 2 = 7.

So, the number of favourable outcomes to the E₁ = 7.

∴ P(E₁) = P(a white or blue ball) = $\dfrac{7}{18}.$

Hence, the probability that the ball drawn is white or blue is $\dfrac{7}{18}.$

(ii) Let E₂ be the event 'a red or black ball is drawn'.

The number of red or black balls = 7 + 4 = 11.

So, the number of favourable outcomes to the E₂ = 11.

∴ P(E₂) = P(a red or black ball) = $\dfrac{11}{18}.$

Hence, the probability that the ball drawn is red or black is $\dfrac{11}{18}.$

(iii) Let E₃ be the event 'not white ball is drawn'.

Probability of not white ball drawn = Probability of any other ball drawn.

No. of balls except white balls = 18 - 5 = 13.

So, the number of favourable outcomes to the E₃ = 13.

∴ P(E₃) = P(not a white ball) = $\dfrac{13}{18}.$

Hence, the probability that the ball drawn is not white is $\dfrac{13}{18}.$

(iv) Let E₄ be the event 'neither white nor black ball is drawn'.

Probability that neither white nor black ball is drawn = Probability of a red or blue ball drawn.

No. of red or blue balls = 7 + 2 = 9.

So, the number of favourable outcomes to the E₄ = 9.

∴ P(E₄) = $\dfrac{9}{18} = \dfrac{1}{2}.$

Hence, the probability that the ball drawn is neither white nor black is $\dfrac{1}{2}.$

A box contains 20 balls bearing numbers 1, 2, 3, 4, ......, 20. A ball is drawn at random from the box. What is the probability that the number on the ball is

(i) an odd number

(ii) divisible by 2 or 3

(iii) prime number

(iv) not divisible by 10?

Answer

A ball is drawn at random from the box it means that all outcomes are equally likely.

Sample space = {1, 2, 3, ......, 20}, which has 20 equally likely outcomes.

(i) Let A be the event 'the number on the ball is odd', then

A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}.

∴ The number of favourable outcomes to the event A = 10.

∴ P(odd number) = $\dfrac{10}{20} = \dfrac{1}{2}.$

Hence, the probability that the ball drawn has an odd number is $\dfrac{1}{2}.$

(ii) Let B be the event 'the number on the ball is divisible by 2 or 3', then

B = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20}.

∴ The number of favourable outcomes to the event B = 13.

∴ P(number is divisible by 2 or 3) = $\dfrac{13}{20}.$

Hence, the probability that the ball drawn has a number that is divisible by 2 or 3 is $\dfrac{13}{20}.$

(iii) Let C be the event 'the number on the ball is prime', then

C = {2, 3, 5, 7, 11, 13, 17, 19}.

∴ The number of favourable outcomes to the event C = 8.

∴ P(prime number) = $\dfrac{8}{20} = \dfrac{2}{5}.$

Hence, the probability that the ball drawn has a prime number is $\dfrac{2}{5}.$

(iv) Let D be the event 'the number on the ball is not divisible by 10', then

D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19}.

∴ The number of favourable outcomes to the event D = 18.

∴ P(number is not divisible by 10) = $\dfrac{18}{20} = \dfrac{9}{10}.$

Hence, the probability that the ball drawn has a number that is not divisible by 10 is $\dfrac{9}{10}.$

Find the probability that a number selected at random from the numbers 1, 2, 3, ......, 35 is a

(i) prime number

(ii) multiple of 7

(iii) multiple of 3 or 5

Answer

A number is selected at random it means that all outcomes are equally likely.

Sample space = {1, 2, 3, ......, 35}, which has 35 equally likely outcomes.

(i) Let A be the event 'the number is prime', then

A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}.

∴ The number of favourable outcomes to the event A = 11.

∴ P(prime number) = $\dfrac{11}{35}.$

Hence, the probability that the number selected is a prime number is $\dfrac{11}{35}.$

(ii) Let B be the event 'the number is multiple of 7', then

B = {7, 14, 21, 28, 35}.

∴ The number of favourable outcomes to the event B = 5.

∴ P(multiple of 7) = $\dfrac{5}{35} = \dfrac{1}{7}.$

Hence, the probability that the number selected is a multiple of 7 is $\dfrac{1}{7}.$

(iii) Let C be the event 'the number is multiple of 3 or 5', then

C = {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35}.

∴ The number of favourable outcomes to the event C = 16.

∴ P(multiple of 3 or 5) = $\dfrac{16}{35}.$

Hence, the probability that the number selected is a multiple of 3 or 5 is $\dfrac{16}{35}.$

Cards marked with numbers 13, 14, 15, ...., 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is

(i) divisible by 5

(ii) a number which is a perfect square.

Answer

A card is selected at random it means that all outcomes are equally likely.

Sample space = {13, 14, 15, ......, 60}, which has 48 equally likely outcomes.

(i) Let A be the event 'the number is divisible by 5', then

A = {15, 20, 25, 30, 35, 40, 45, 50, 55, 60}.

∴ The number of favourable outcomes to the event A = 10.

∴ P(divisible by 5) = $\dfrac{10}{48} = \dfrac{5}{24}.$

Hence, the probability that the number selected is divisible by 5 is $\dfrac{5}{24}.$

(ii) Let B be the event 'the number is a perfect square', then

B = {16, 25, 36, 49}.

∴ The number of favourable outcomes to the event B = 4.

∴ P(perfect square) = $\dfrac{4}{48} = \dfrac{1}{12}.$

Hence, the probability that the number selected is a perfect square is $\dfrac{1}{12}.$

A box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn at random from the box. Find the probability that the card drawn from the box has

(i) an odd number

(ii) a perfect square number.

Answer

A card is selected at random it means that all outcomes are equally likely.

Sample space = {14, 15, 16, ......, 99}, which has 86 equally likely outcomes.

(i) Let A be the event 'odd number', then

A = {15, 17, 19, 21, 23, ........., 91, 93, 95, 97, 99}.

∴ The number of favourable outcomes to the event A = 43.

∴ P(odd number) = $\dfrac{43}{86} = \dfrac{1}{2}.$

Hence, the probability that the number selected is an odd number is $\dfrac{1}{2}.$

(ii) Let B be the event 'a perfect square number', then

B = {16, 25, 36, 49, 64, 81}.

∴ The number of favourable outcomes to the event B = 6.

∴ P(a perfect square number) = $\dfrac{6}{86} = \dfrac{3}{43}.$