Probability

A coin is tossed once.

(i) Describe the sample space S.

(ii) Find the probability of getting a tail.

Answer

(i) When a coin is tossed, we get either Head (H) or Tail (T).

∴ Sample space (S) = {H, T}

⇒ n(S) = 2

Hence, sample space = {H, T}.

(ii) Let E be the event of getting a tail.

Then, E = {T}

⇒ n(E) = 1

∴ P(getting a tail) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{2}$

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

Two coins are tossed simultaneously. Describe the sample space S. Find the probability of getting:

(i) two heads

(ii) at least one head

(iii) at most one head

(iv) exactly one head

(v) no head

Answer

When you toss two coins simultaneously, we get either both heads, first head second tail, first tail second head, both tails.

S = {HH, HT, TH, TT}

Total number of outcomes = 4

(i) Two heads

Number of favorable outcomes (two heads) = 1 (HH)

∴ P(getting two heads) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{4}$

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

(ii) At least one head

Number of favorable outcomes (Getting at least one head) = 3 (HT, TH and HH)

∴ P(getting at least one head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{4}$

Hence, the probability of getting at least one head is $\dfrac{3}{4}$.

(iii) At most one head

Number of favorable outcomes (Getting at most one head) = 3 (HT, TH and TT)

∴ P(getting at most one head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{4}$

Hence, the probability of getting at most one head is $\dfrac{3}{4}$.

(iv) Exactly one head

Number of favorable outcomes (Getting exactly one head) = 2 (HT, TH)

∴ P(getting exactly one head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{4} = \dfrac{1}{2}.$

Hence, the probability of getting exactly one head is $\dfrac{1}{2}$.

(v) No head

Number of favorable outcomes (Getting no head) = 1 (TT)

∴ P(getting no head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{4}.$

Hence, the probability of getting no head is $\dfrac{1}{4}$.

Three coins are tossed simultaneously. Describe the sample space S. Find the probability of getting:

(i) at most 2 heads

(ii) at least 2 heads

(iii) exactly 2 heads

Answer

When you toss three coins simultaneously.

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Total number of outcomes = 8

(i) at most 2 heads

Number of favorable outcomes (Getting at most 2 heads) = 7 (HHT, HTH, THH, HTT, THT, TTH, TTT)

∴ P(getting at most 2 heads) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{8}.$

Hence, the probability of getting at most 2 heads is $\dfrac{7}{8}$.

(ii) at least 2 heads

Number of favorable outcomes (Getting at least 2 heads) = 4 (HHT, HTH, THH, HHH)

∴ P(getting at least 2 heads) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{8} = \dfrac{1}{2}.$

Hence, the probability of getting at least 2 heads is $\dfrac{1}{2}$.

(iii) exactly 2 heads

Number of favorable outcomes (Getting exactly 2 heads) = 3 (HHT, HTH, THH)

∴ P(getting exactly 2 heads) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{8} .$

Hence, the probability of getting exactly 2 heads is $\dfrac{3}{8}$.

A die is thrown once. What is the probability of getting:

(i) an odd number

(ii) a number greater than 4

(iii) a number less than 5

(iv) the number 5

Answer

In a single throw of die,

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

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

A = {1, 3, 5}.

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

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{6} = \dfrac{1}{2} .$

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

(ii) Let B be the event of getting a number greater than 4, then

B = {5, 6}.

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

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{6} = \dfrac{1}{3} .$

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

(iii) Let C be the event of getting a number less than 5, then

C = {1, 2, 3, 4}.

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

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{6} = \dfrac{2}{3} .$

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

(iv) Let D be the event of getting the number 5, then

D = {5}.

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

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{6}.$

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

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

(i) 10 as the sum of two numbers that turn up

(ii) a doublet of even numbers

(iii) a total of at least 10

(iv) a multiple of 3 as the sum of two numbers that turn up

Answer

(i) Let A be the event of 10 as the sum of two numbers that turn up, then

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

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

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{36} = \dfrac{1}{12}.$

Hence, the probability of 10 as the sum of two numbers that turn up is $\dfrac{1}{12}.$

(ii) Let B be the event of a doublet of even numbers, then

B = {(2, 2), (4, 4), (6, 6)}.

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

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{36} = \dfrac{1}{12}.$

Hence, the probability of getting a doublet of even numbers is $\dfrac{1}{12}.$

(iii) Let C be the event of a total of at least 10, then

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

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

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}.$

Hence, the probability of getting a total of at least 10 is $\dfrac{1}{6}.$

(iv) Let D be the event of a multiple of 3 as the sum of two numbers that turn up , then

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

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

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{36} = \dfrac{1}{3}.$

Hence, the probability of getting a multiple of 3 as the sum of two numbers that turn up is $\dfrac{1}{3}.$

A box of 160 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from the box. What is the probability that the bulb drawn is:

(i) defective?

(ii) non-defective?

Answer

Given,

Total number of bulbs = 160

Number of defective bulbs = 12

Number of non defective bulbs = 160 - 12 = 148

(i) Let A be the event of taking out defective bulb , then

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

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{160} = \dfrac{3}{40}.$

Hence, the probability of taking out a defective bulb is $\dfrac{3}{40}.$

(ii) Let B be the event of taking out non-defective bulb , then

∴ The number of favourable outcomes to the event B = 148.

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{148}{160} = \dfrac{37}{40}.$

Hence, the probability of taking out a non-defective bulb is $\dfrac{37}{40}.$

A box contains 16 cards bearing numbers 1, 2, 3, 4, …, 15, 16 respectively. A card is drawn at random from the box. What is the probability that the number on the card is:

(i) an odd number?

(ii) a prime number?

(iii) a number divisible by 3?

(iv) a number not divisible by 4?

Answer

Given,

Total number of outcomes = 16

(i) Let A be the event of drawing an odd number , then

A = {1, 3, 5, 7, 9, 11, 13, 15}

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

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{8}{16} = \dfrac{1}{2}.$

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

(ii) Let B be the event of drawing a prime number, then

B = {2, 3, 5, 7, 11, 13}

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

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{16} = \dfrac{3}{8}.$

Hence, the probability of drawing a prime number is $\dfrac{3}{8}.$

(iii) Let C be the event of drawing a number divisible by 3, then

C = {3, 6, 9, 12, 15}

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

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{16}$

Hence, the probability of drawing a number divisible by 3 is $\dfrac{5}{16}.$

(iv) Let D be the event of drawing a number not divisible by 4, then

D = {1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15}

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

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{16} = \dfrac{3}{4}$

Hence, the probability of drawing a number not divisible by 4 is $\dfrac{3}{4}.$

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

Given,

Cards in the bag:

Sample space = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}.

Total number of outcomes = 10

(i) Let A be the event of drawing a card with prime number, then

A = {2}

∴ The number of favourable outcomes to the event A = 1

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{10}$

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

(ii) Let B be the event of drawing a card with number divisible by 4, then

B = {4, 8, 12, 16, 20}

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

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{10} = \dfrac{1}{2}$

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

(iii) Let C be the event of drawing a card with multiple of 6, then

C = {6, 12, 18}

∴ The number of favourable outcomes to the event C = 3

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{10}$

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

(iv) Let D be the event of drawing a card with an odd number, then

D = ∅

∴ The number of favourable outcomes to the event D = 0

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{0}{10} = 0$

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

A box contains 15 balls bearing numbers 1, 2, 3, …, 14, 15 respectively. A ball is drawn at random from the box. Find the probability that the number on the ball is:

(i) an even number

(ii) a number divisible by 5

(iii) the number 6

(iv) a number lying between 8 and 12

(v) a number greater than 9

(vi) a number less than 6

Answer

Given,

Balls in the box are numbered

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

Total number of outcomes = 15

(i) Let A be the event of getting an even number, then

A = {2, 4, 6, 8, 10, 12, 14}

∴ The number of favourable outcomes to the event A = 7

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{15}$

Hence, the probability of getting an even number is $\dfrac{7}{15}$.

(ii) Let B be the event of getting a number divisible by 5, then

B = {5, 10, 15}

∴ The number of favourable outcomes to the event B = 3

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{15} = \dfrac{1}{5}$

Hence, the probability of getting a number divisible by 5 is $\dfrac{1}{5}$.

(iii) Let C be the event of getting the number 6, then

C = {6}

∴ The number of favourable outcomes to the event C = 1

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{15}$

Hence, the probability of getting the number 6 is $\dfrac{1}{15}$.

(iv) Let D be the event of getting a number between 8 and 12, then

D = {9, 10, 11}

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

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{15} = \dfrac{1}{5}$

Hence, the probability of getting a number lying between 8 and 12 is $\dfrac{1}{5}$.

(v) Let E be the event of getting a number greater than 9, then

E = {10, 11, 12, 13, 14, 15}

∴ The number of favourable outcomes to the event E = 6

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{15} = \dfrac{2}{5}$

Hence, the probability of getting a number greater than 9 is $\dfrac{2}{5}$.

(vi) Let F be the event of getting a number less than 6, then

F = {1, 2, 3, 4, 5}

∴ The number of favourable outcomes to the event F = 5

∴ P(F) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{15} = \dfrac{1}{3}$

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

There are 25 discs numbered 1 to 25. They are put in a closed box and shaken thoroughly. 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

Given,

There are 25 discs numbered from 1 to 25.

Sample space:S={1, 2, 3, …, 25}

Total number of outcomes = 25

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

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

∴ The number of favourable outcomes to the event A = 13

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{13}{25}$

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

(ii) Let B be the event of getting a number divisible by both 2 and 3, then

B = {6, 12, 18, 24}

∴ The number of favourable outcomes to the event B = 4

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{25}$

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

(iii) Let C be the event of getting a number less than 16, then

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

∴ The number of favourable outcomes to the event C = 15

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{15}{25} = \dfrac{3}{5}$

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

In a class of 40 students, there are 16 boys and the rest are girls. From these students, one is selected at random. What is the probability that the selected student is a girl?

Answer

Given,

Total number of students = 40

Number of boys = 16

Let G be the event of selecting a girl,

∴ The number of favourable outcomes to the event G = 24

∴ P(G) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{24}{40} = \dfrac{3}{5}$

Hence, the probability of selecting a girl is $\dfrac{3}{5}$.

A bag contains 8 red, 4 white and 3 black balls. One ball is drawn at random. What is the probability that the ball drawn is:

(i) white?

(ii) red or white?

(iii) neither red nor white?

(iv) not red?

Answer

Given,

Total number of outcomes = 15

(i) Let A be the event of getting white ball, then

∴ The number of favourable outcomes to the event A = 4

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{15}$

Hence, the probability of getting white ball is $\dfrac{4}{15}$.

(ii) Let B be the event of getting white or red ball, then

∴ The number of favourable outcomes to the event B = 12(white + red balls)

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{15}= \dfrac{4}{5}$

Hence, the probability of getting white or red ball is $\dfrac{4}{5}$.

(iii) Let C be the event of getting neither red nor white ball, then

∴ The number of favourable outcomes to the event C = 3 (black balls)

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{15}= \dfrac{1}{5}$

Hence, the probability of getting neither red nor white ball is $\dfrac{1}{5}$.

(iv) Let D be the event of not getting red ball, then

∴ The number of favourable outcomes to the event D = 7(white + black balls)

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{15}$

Hence, the probability of not getting red ball is $\dfrac{7}{15}$.

A bag contains 6 black, 5 white and 9 green balls. One ball is drawn at random. What is the probability that the ball drawn is:

(i) black?

(ii) not green?

(iii) either white or green?

(iv) neither white nor black?

Answer

Given,

Total number of outcomes = 6 (Black) + 5 (White) + 9 (Green) = 20

(i) Let A be the event of getting black ball, then

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

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{20} = \dfrac{3}{10}$

Hence, the probability of getting black ball is $\dfrac{3}{10}$.

(ii) Let B be the event of not getting green ball, then

∴ The number of favourable outcomes to the event B = 6 (Black) + 5 (White) = 11

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{11}{20}$

Hence, the probability of not getting green ball is $\dfrac{11}{20}$.

(iii) Let C be the event of getting either white or green ball, then

∴ The number of favourable outcomes to the event C = 5 (White) + 9 (Green) = 14

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{14}{20}= \dfrac{7}{10}$

Hence, the probability of getting either white or green ball is $\dfrac{7}{10}$.

(iv) Let D be the event of getting neither white nor black ball , then

∴ The number of favourable outcomes to the event D = 9(green balls)

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{9}{20}$

Hence, the probability of getting neither white nor black ball is $\dfrac{9}{20}$.

One card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of drawing:

(i) an ace

(ii) a 5 of a red suit

(iii) a black queen

(iv) a jack of spades

(v) a 10 of hearts

(vi) a face card

Answer

Given,

Total number of outcomes = 52

(i) Let A be the event of getting an ace, then

∴ The number of favourable outcomes to the event A = 4

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{52} = \dfrac{1}{13}$

Hence, the probability of getting an ace is $\dfrac{1}{13}$.

(ii) Let B be the event of getting 5 of red suit, then

∴ The number of favourable outcomes to the event B = 2

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}$

Hence, the probability of getting 5 of red suit is $\dfrac{1}{26}$.

(iii) Let C be the event of getting a black queen, then

∴ The number of favourable outcomes to the event C = 2 (one of club and one of spade)

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}$

Hence, the probability of getting a black queen is $\dfrac{1}{26}$.

(iv) Let D be the event of getting a jack of spades, then

∴ The number of favourable outcomes to the event D = 1

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{52}$

Hence, the probability of getting a jack of spades is $\dfrac{1}{52}$.

(v) Let E be the event of getting a 10 of hearts, then

∴ The number of favourable outcomes to the event E = 1

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{52}$

Hence, the probability of getting a 10 of hearts is $\dfrac{1}{52}$.

(vi) Let F be the event of getting a face card, then

There are 12 face cards in a deck.

∴ The number of favourable outcomes to the event F = 12

∴ P(F) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{52} = \dfrac{3}{13}$.

Hence, the probability of getting a face card is $\dfrac{3}{13}$.

A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that the card drawn is:

(i) either a king or a queen

(ii) neither a king nor a queen

Answer

Given,

Total number of outcomes = 52

(i) Let A be the event of getting either a king or a queen, then

∴ The number of favourable outcomes to the event A = 4 (Kings) + 4 (Queens) = 8

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{8}{52} = \dfrac{2}{13}$

Hence, the probability of getting either a king or a queen is $\dfrac{2}{13}$.

(ii) Let B be the event of getting neither a king nor a queen, then

∴ The number of favourable outcomes to the event B = Total cards - (Kings + Queens) = 44

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{44}{52} = \dfrac{11}{13}$

Hence, the probability of getting neither a king nor a queen is $\dfrac{11}{13}$.

If the probability of winning a game is 0.6, what is the probability of losing the game?

Answer

We know that,

P(E) + P(not E) = 1

Given,

Probability of winning game = 0.6

P(winning) + P(losing) = 1

P(losing) = 1 - 0.6 = 0.4

Hence, the probability of losing game is 0.4

Fill in the blanks:

(i) The probability of a sure event is __________.

(ii) The probability of an impossible event is __________.

(iii) For an event E, we have P(E) + P(not E) = __________.

(iv) For an event E, we have () ≤ P(E) ≤ ().

Answer

Fill in the blanks:

(i) The probability of a sure event is 1.

(ii) The probability of an impossible event is 0.

(iii) For an event E, we have P(E) + P(not E) = 1.

(iv) For an event E, we have 0 ≤ P(E) ≤ 1.

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

Given,

Sample space S = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p}

Total number of outcomes = 16

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

A = {a, e, i, o}

∴ The number of favourable outcomes to the event A = 4

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{16} = \dfrac{1}{4}$

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

(ii) Let B be the event of getting a consonant, then

B = {b, c, d, f, g, h, j, k, l, m, n, p}

∴ The number of favourable outcomes to the event B = 12

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{16} = \dfrac{3}{4}$

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

(iii) Let C be the event of not getting letters of word median, then

C = {b, c, f, g, h, j, k, l, o, p}

∴ The number of favourable outcomes to the event C = 10

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{10}{16} = \dfrac{5}{8}$

Hence, the probability of not getting letters of word median is $\dfrac{5}{8}$.

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,

(i) MONDAY?

(ii) which does not appear in MONDAY?

(iii) which appears both in SUNDAY and MONDAY?

Answer

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

No. of cards = 8

(i) 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 = $\dfrac{3}{4}$.

(ii) 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 = $\dfrac{1}{4}$.

(iii) 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 = $\dfrac{1}{2}$.

A bag contains 25 cards, numbered through 1 to 25. A card is drawn at random. What is the probability that the number on the card drawn is:

(i) a multiple of 5

(ii) a perfect square

(iii) a prime number

Answer

Given,

Sample space = {1, 2, 3, 4, 5, ....., 25}

Total number of outcomes = 25

(i) Let A be the event of getting a multiple of 5, then

A = {5, 10, 15, 20, 25}

∴ The number of favourable outcomes to the event A = 5

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$

Hence, the probability of getting a multiple of 5 is $\dfrac{1}{5}$.

(ii) Let B be the event of getting a perfect square, then

B = {1, 4, 9, 16, 25}

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

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$

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

(iii) Let C be the event of getting a prime number, then

C = {2, 3, 5, 7, 11, 13, 17, 19, 23}

∴ The number of favourable outcomes to the event C = 9

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{9}{25}$

Hence, the probability of getting a prime number is $\dfrac{9}{25}$.

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 a red ball?

Answer

Total number of outcomes = 5 white + 2 red + 3 black balls = 10

Let A be the event of getting a red ball.

The number of favorable outcomes to the event A = 2

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{10} = \dfrac{1}{5}$

Hence, the probability of drawing a red ball is $\dfrac{1}{5}$.

A letter of the word SECONDARY is selected at random. What is the probability that the letter selected is not a vowel?

Answer

Given, Word = SECONDARY The letters are {S, E, C, O, N, D, A, R, Y}

Total number of outcomes = 9

The letters that are consonants = {S, C, N, D, R, Y}.

Let B be the event of selecting a letter that is not a vowel.

The number of favorable outcomes to the event B = 6

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{9} = \dfrac{2}{3}$

Hence, the probability that the letter selected is not a vowel is $\dfrac{2}{3}$.

Ms. Sushmita went to a fair and participated in a game. The game consisted of a box having number cards with numbers from 01 to 30. The three prizes were as per the given table:

Find the probability of winning a:

(i) Wall Clock

(ii) Water Bottle

(iii) Purse

Answer

Given,

Total number of outcomes = 30

(i) Given,

Numbers that are perfect squares (between 1 to 30) are 1, 4, 9, 16, 25 (5 numbers).

∴ No. of favourable outcomes = 5 perfect squares

P(perfect square) $=\dfrac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{30} = \dfrac{1}{6}$.

Hence, probability of numbers that are perfect squares = $\dfrac{1}{6}$.

(ii) Given,

Even numbers that are also multiples of 3 (between 1 to 30) are 6, 12, 18, 24, 30 (5 numbers)

∴ No. of favourable outcomes = 5 Even numbers that are also multiples of 3

P(even and multiple of 3) $=\dfrac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{30} = \dfrac{1}{6}$

Hence, probability of numbers that are multiples of 3 and even number = $\dfrac{1}{6}$.

(iii) Given,

Prime numbers (1 to 30) are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 (10 numbers)

∴ No. of favourable outcomes = 10 Prime numbers

P(prime)

$=\dfrac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \dfrac{10}{30} = \dfrac{1}{3}$

Hence, probability of prime numbers = $\dfrac{1}{3}$.

A box containing cards numbered between 0 and 100 are shuffled and a card is picked at random. Find the probability of getting a card which is:

(i) divisible by 6.

(ii) not divisible by 6.

Answer

The cards are numbered from 0 to 100.

Total number of cards = 101 (as there is no. zero card also)

(i) Favorable outcome = Getting a card having a number which is divisible by 6.

{0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96}

Probability $=\dfrac{\text{No of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{17}{101}$.

Hence, probability of getting a card which is divisible by 6 = $\dfrac{17}{101}$.

(ii) Favorable outcome = Getting a card having a number which is not divisible by 6.

P(getting a card not divisible by 6) = 1 - P(getting a card divisible by 6)

= $1 - \dfrac{17}{101}$

= $\dfrac{101 - 17}{101}$

= $\dfrac{84}{101}$.

Hence, probability of getting a card which is not divisible by 6 = $\dfrac{84}{101}$.

There are some red, green and white marbles in a box. One marble is picked up at random from this box. If the probability of picking up a red marble is $\dfrac{2}{9}$ and that of picking up a green marble is $\dfrac{4}{9}$ then find the :

(i) probability of picking up a white marble.

(ii) number of green marbles, if total number of marbles is 54.

(iii) probability of not picking up a red marble.

Answer

(i) The sum of the probabilities of all possible outcomes is equal to 1.

The possible outcomes are picking a red, green, or white marble.

P(red) + P(green) + P(white) = 1

Given,

P(red) = $\dfrac{2}{9}$

P(green) = $\dfrac{4}{9}$

$\Rightarrow \dfrac{2}{9} + \dfrac{4}{9} + P(\text{white}) = 1 \\[1em] \Rightarrow \dfrac{6}{9} + P(\text{white}) = 1 \\[1em] \Rightarrow P(\text{white}) = 1 - \dfrac{6}{9} \\[1em] \Rightarrow P(\text{white}) = \dfrac{9 - 6}{9} \\[1em] \Rightarrow P(\text{white}) = \dfrac{3}{9} = \dfrac{1}{3}.$

Hence, probability of picking up a white marble = $\dfrac{1}{3}$.

(ii) Given,

Total number of marbles = 54

P(green) $=\dfrac{\text{No of favorable outcomes}}{\text{Total number of outcomes}}$.

$\Rightarrow \dfrac{4}{9} = \dfrac{\text{Number of green marbles}}{54} \\[1em] \Rightarrow \dfrac{4}{9} \times 54 = \text{Number of green marbles} \\[1em] \Rightarrow 4 \times 6 = \text{Number of green marbles} \\[1em] \Rightarrow \text{Number of green marbles} = 24.$

Hence, number of green marbles is 24.

(iii) Given,

P(red) = $\dfrac{2}{9}$

P(not picking a red) = 1 - P(red)

= 1 - $\dfrac{2}{9}$

= $\dfrac{9 - 2}{9}$

= $\dfrac{7}{9}$.

Hence, probability of not picking up a red marble is $\dfrac{7}{9}$.

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

1. $\Big(\dfrac{3}{5}\Big)$

2. 25%

3. 0.96

4. −0.5

Answer

Probability is a measure of the likelihood of an event occurring, and it must always fall within a specific range:

0 ≤ P(E) ≤ 1

Probability can't be negative.

Hence, option 4 is the correct option.

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

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

2. 0.25

3. $\dfrac{1}{33}$

4. 67%.

Answer

Probability of any event is always between 0 and 1.

$\dfrac{5}{4}$ = 1.25, which is greater than 1, which is not possible.

Hence, option 1 is the correct option.

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

1. $\Big(\dfrac{1.8}{3}\Big)$

2. $\Big(\dfrac{1}{0.4}\Big)$

3. $\Big(\dfrac{0.4}{5}\Big)$

4. $\Big(\dfrac{2}{65}\Big)$

Answer

We know that,

0 ≤ P(E) ≤ 1

$\Big(\dfrac{1}{0.4}\Big)$ = 2.5

Since this value is greater than 1, it cannot be a probability.

Hence, option 2 is the correct option.

When polynomial x³ - 3x² - 6x + 8 is divided by (x + 2), the remainder is zero. The probability of (x + 2) to be one of the factors of the given polynomial is:

1. 0

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

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

4. 1

Answer

Since, on dividing x³ - 3x² - 6x + 8 by (x + 2), the remainder is zero.

Thus, (x + 2) is the polynomial of x³ - 3x² - 6x + 8.

∴ The probability of (x + 2) to be one of the factors of the polynomial = 1.

Hence, option 4 is the correct option.

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

1. $\Big(\dfrac{1}{6}\Big)$

2. $\Big(\dfrac{1}{3}\Big)$

3. $\Big(\dfrac{1}{2}\Big)$

4. $\Big(\dfrac{2}{3}\Big)$

Answer

When a die is thrown, the possible outcomes are:

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

Total number of outcomes = 6

Let E be the event of getting a number divisible by 3, then

E = {3, 6}

The number of favorable outcomes to the event E = 2

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}$

Hence, option 2 is the correct option.

If the probability of an event is p, then which of the following holds true?

1. −1 ≤ p ≤ 1

2. 0 ≤ p ≤ ∞

3. 0 ≤ p ≤ 1

4. −∞ < p < ∞

Answer

Probability is a measure of the likelihood of an event occurring, and it must always fall within a specific range:

0 ≤ p ≤ 1

Hence, option 3 is the correct option.

In a single throw of a die, the probability of getting a number greater than 4 is:

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{1}{3}\Big)$

3. $\Big(\dfrac{1}{4}\Big)$

4. $\Big(\dfrac{2}{3}\Big)$

Answer

When a die is thrown, there are 6 possible outcomes:

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

Total number of outcomes = 6

Let E be the event of getting a number greater than 4, then

E = {5, 6}

The number of favorable outcomes to the event E = 2

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}$

Hence, option 2 is the correct option.

In a single throw of a die, the probability of getting a prime number, is:

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{1}{3}\Big)$

3. $\Big(\dfrac{1}{4}\Big)$

4. $\Big(\dfrac{2}{3}\Big)$

Answer

When a die is thrown, there are 6 possible outcomes:

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

Total number of outcomes = 6

Let E be the event of getting a prime number, then

E = {2, 3, 5}

The number of favorable outcomes to the event E = 3

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$

Hence, option 1 is the correct option.

A die is thrown once. The probability of getting a number which has at least two factors is:

1. 1

2. $\Big(\dfrac{1}{2}\Big)$

3. $\Big(\dfrac{2}{3}\Big)$

4. $\Big(\dfrac{5}{6}\Big)$

Answer

When a die is thrown, there are 6 possible outcomes:

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

Total number of outcomes = 6

Let E be the event of getting a number with at least two factors, then

E = {2, 3, 4, 5, 6}

The number of favorable outcomes to the event E = 5

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{6}$

Hence, option 4 is the correct option.

There are 3 yellow, 5 white and 7 green balls in a bag. A ball is drawn from the bag at random. The probability that the ball drawn is not white, is:

1. $\Big(\dfrac{1}{3}\Big)$

2. $\Big(\dfrac{2}{3}\Big)$

3. $\Big(\dfrac{3}{5}\Big)$

4. $\Big(\dfrac{1}{2}\Big)$

Answer

Total number of outcomes = 3 yellow + 5 white + 7 green balls = 15

Let E be the event of not getting white balls.

The number of favorable outcomes to the event E = 3 yellow balls + 7 green balls = 10

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{10}{15} = \dfrac{2}{3}$

Hence, option 2 is the correct option.

In the above question, what is the probability that a ball drawn at random from the bag is not black?

1. 0

2. $\Big(\dfrac{1}{2}\Big)$

3. 1

4. cannot be computed

Answer

Total number of outcomes = 3 yellow + 5 white + 7 green balls = 15

Let E be the event of not getting black balls.

The number of favorable outcomes to the event E = 15

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{15}{15} = 1$

Hence, option 3 is the correct option.

Cards marked with numbers 1 to 100 are placed in a box and mixed thoroughly. A card is drawn at random from the box. The probability that the selected card bears a perfect square number is:

1. $\Big(\dfrac{1}{10}\Big)$

2. $\Big(\dfrac{2}{25}\Big)$

3. $\Big(\dfrac{9}{100}\Big)$

4. $\Big(\dfrac{11}{100}\Big)$

Answer

The cards are numbered from 1 to 100.

Total number of outcomes = 100

Let E be the event of getting perfect square, then

E = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

The number of favorable outcomes to the event E = 10

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{10}{100} = \dfrac{1}{10}$

Hence, option 1 is the correct option.

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3?

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{2}{5}\Big)$

3. $\Big(\dfrac{3}{10}\Big)$

4. $\Big(\dfrac{3}{20}\Big)$

Answer

The tickets are numbered from 1 to 20.

Total number of outcomes = 20

Let E be the event of getting a number multiple of 3, then

E = {3, 6, 9, 12, 15, 18}

The number of favorable outcomes to the event E = 6

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{20} = \dfrac{3}{10}$

Hence, option 3 is the correct option.

Cards numbered 1 to 20 are placed in a box and mixed thoroughly. A card is drawn at random from the box. What is the probability that the card drawn bears a number which is a multiple of 3 or 5 or both?

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{2}{5}\Big)$

3. $\Big(\dfrac{8}{15}\Big)$

4. $\Big(\dfrac{9}{20}\Big)$

Answer

The cards are numbered from 1 to 20.

Total number of outcomes = 20

Let E be the event of getting a number multiple of 3 or 5, then

E = {3, 5, 6, 9, 10, 12, 15, 18, 20}

The number of favorable outcomes to the event E = 9

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{9}{20}$

Hence, option 4 is the correct option.

A number is chosen at random from the numbers −4, −3, −2, −1, 0, 1, 2, 3, 4. What is the probability that the square of this number is less than or equal to 2?

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{1}{3}\Big)$

3. $\Big(\dfrac{4}{9}\Big)$

4. $\Big(\dfrac{5}{9}\Big)$

Answer

Sample space = {−4, −3, −2, −1, 0, 1, 2, 3, 4}

Total number of outcomes = 9

Let E be the event of choosing the number whose square is less than or equal to 2, then

E = {-1, 0, 1}

The number of favorable outcomes to the event E = 3

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{9} = \dfrac{1}{3}$

Hence, option 2 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

Given,

Number of red marbles = 3

Number of blue marbles = 2

The total number of marbles in the bag is 3 + 2 = 5.

Number of black marbles in the bag = 0

Probability = $\dfrac{\text{No of favorable outcomes}}{\text{Total number of outcomes}}$.

= $\dfrac{0}{5}$

= 0.

Hence, option 1 is the correct option.

Choose the incorrect statement.

If a card is picked at random from cards numbered 1 to 16, then:

1. the probability that the drawn card bears a number which is a factor of 16 is $\Big(\dfrac{1}{4}\Big)$

2. the probability that the drawn card bears an odd composite number is $\Big(\dfrac{1}{8}\Big)$

3. the probability that the drawn card bears a number which is a multiple of both 2 and 3 is $\Big(\dfrac{1}{8}\Big)$

4. the probability that the drawn card bears a number which is both a perfect square and a perfect cube is $\Big(\dfrac{1}{16}\Big)$

Answer

S = {1, 2, 3, ....., 16}, where the total number of outcomes is 16.

Let E be the event of getting a card with number that divide 16 perfectly, then

E = {1, 2, 4, 8, 16}

The number of favorable outcomes to the event E = 5

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{16}$

The probability that the drawn card bears a number which is a factor of 16 is $\Big(\dfrac{1}{4}\Big)$ is incorrect.

Hence, option 1 is the correct option.

A letter of English alphabet is chosen at random. The probability that it is a letter of the word ENGINEERING is:

1. $\Big(\dfrac{5}{26}\Big)$

2. $\Big(\dfrac{3}{13}\Big)$

3. $\Big(\dfrac{11}{26}\Big)$

4. $\Big(\dfrac{6}{13}\Big)$

Answer

The experiment consists of choosing a letter from the entire English alphabet.

Total number of outcomes = 26

Let E be the event of choosing a letter of the word ENGINEERING, then

E = {E, N, G, I, R}

The number of favorable outcomes to the event E = 5

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{26}$

Hence, option 1 is the correct option.

A letter is picked at random from the letters of the word MATRICULATION. The probability that the chosen letter is a vowel, is:

1. $\Big(\dfrac{2}{5}\Big)$

2. $\Big(\dfrac{1}{2}\Big)$

3. $\Big(\dfrac{6}{13}\Big)$

4. $\Big(\dfrac{4}{13}\Big)$

Answer

The total number of letters in the word MATRICULATION = 13

Let E be the event of choosing a vowel from word MATRICULATION, then

E = {A, I, U, A, I, O}

The number of favorable outcomes to the event E = 6

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{13}$

Hence, option 3 is the correct option.

In a family of 3 children, the probability of having at least one boy is:

1. $\Big(\dfrac{1}{8}\Big)$

2. $\Big(\dfrac{5}{8}\Big)$

3. $\Big(\dfrac{3}{4}\Big)$

4. $\Big(\dfrac{7}{8}\Big)$

Answer

The sample space is: {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}

The total number of possible combinations = 8

Let E be the event of having at least one boy, then

E = {BBB, BBG, BGB, GBB, BGG, GBG, GGB}

The number of favorable outcomes to the event E = 7

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{8}$

Hence, option 4 is the correct option.

A number is chosen randomly from 10 to 99 (both inclusive) such that each number is equally likely to be chosen. The probability that at least one digit of the chosen number is 8 is:

1. $\Big(\dfrac{1}{5}\Big)$

2. $\Big(\dfrac{1}{9}\Big)$

3. $\Big(\dfrac{1}{10}\Big)$

4. $\Big(\dfrac{19}{20}\Big)$

Answer

The numbers are chosen from 10 to 99 inclusive.

Total number of outcomes = 90

Let E be the event of choosing a number with At Least One Digit as 8 , then

E = {18, 28, 38, 48, 58, 68, 78, 88, 98, 80, 81, 82, 83, 84, 85, 86, 87, 89}

The number of favorable outcomes to the event E = 18

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{18}{90} = \dfrac{1}{5}$

Hence, option 1 is the correct option.

A letter is chosen at random from all the letters of the English alphabets. The probability that the letter chosen is a vowel is:

1. $\Big(\dfrac{4}{26}\Big)$

2. $\Big(\dfrac{5}{26}\Big)$

3. $\Big(\dfrac{21}{26}\Big)$

4. $\Big(\dfrac{5}{24}\Big)$

Answer

There are 26 letters in the English alphabet.

Total number of outcomes = 26

Let E be the event of choosing a vowel , then

E = {A, E, I, O, U}

The number of favorable outcomes to the event E = 5

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{26}$

Hence, option 2 is the correct option.

One card is drawn at random from a pack of 52 playing cards. What is the probability that the card drawn is a face card?

1. $\Big(\dfrac{1}{13}\Big)$

2. $\Big(\dfrac{3}{13}\Big)$

3. $\Big(\dfrac{1}{4}\Big)$

4. $\Big(\dfrac{9}{52}\Big)$

Answer

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

Let E be the event of choosing a face card , then

There are 4 suits in a deck, and each suit has 3 face cards.

The number of favorable outcomes to the event E = 4 suits × 3 face cards = 12

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{52} = \dfrac{3}{13}$

Hence, option 2 is the correct option.

A card is drawn from a pack of 52 playing cards. The probability of getting a queen of club or a king of heart is:

1. $\Big(\dfrac{1}{13}\Big)$

2. $\Big(\dfrac{2}{13}\Big)$

3. $\Big(\dfrac{1}{26}\Big)$

4. $\Big(\dfrac{1}{52}\Big)$

Answer

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

Let E be the event of choosing a queen of club or a king of heart , then

There is 1 Queen of Clubs and 1 King of Hearts in the deck.

The number of favorable outcomes to the event E = 2

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}$

Hence, option 3 is the correct option.

One card is drawn at random from a pack of 52 playing cards. The probability that the card drawn is either a red card or a king is:

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{6}{13}\Big)$

3. $\Big(\dfrac{7}{13}\Big)$

4. $\Big(\dfrac{27}{52}\Big)$

Answer

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

There are 26 red cards (13 hearts and 13 diamonds) and 4 kings in a deck, 2 of these kings (King of Hearts and King of Diamonds) are red.

Hence, no. of cards which are red or king = 26 Red Cards + 2 Black Kings = 28

Let E be the event of choosing either a red card or a king , then

The number of favorable outcomes to the event E = 28

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{28}{52} = \dfrac{7}{13}$

Hence, option 3 is the correct option.

When a die is cast with numbering on its faces, as shown, the ratio of the probability of getting a composite number to the probability of getting a prime number is :

1. 2 : 3

2. 3 : 2

3. 1 : 3

4. 1 : 2

Answer

From numbers 1 to 6

Prime numbers = 2, 3, 5.

Composite numbers = 4, 6.

Probability of getting a prime number = $\dfrac{\text{No. of prime numbers}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$.

Probability of getting a composite number

= $\dfrac{\text{No. of composite numbers}}{\text{No. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}$.

Probability of getting a composite number to probability of getting a prime number = $\dfrac{1}{3} : \dfrac{1}{2}$ = 2 : 3.

Hence, option 1 is the correct option.

From a pack of 52 playing cards, one card is drawn at random. What is the probability that the card drawn is a ten or a spade?

1. $\Big(\dfrac{1}{26}\Big)$

2. $\Big(\dfrac{1}{13}\Big)$

3. $\Big(\dfrac{4}{13}\Big)$

4. $\Big(\dfrac{17}{52}\Big)$

Answer

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

There are 13 spade cards and 4 tens in a deck, Ten of Spades is a card that is both a spade and a ten.

Hence, no. of cards that are tens or spades = 13 Spades + 3 Tens

Let E be the event of choosing a ten or a spade , then

The number of favorable outcomes to the event E = 16

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{16}{52} = \dfrac{4}{13}$

Hence, option 3 is the correct option.

A card is drawn at random from a pack of 52 playing cards. The probability of drawing a card which is neither a spade nor a king, is:

1. $\Big(\dfrac{17}{52}\Big)$

2. $\Big(\dfrac{4}{13}\Big)$

3. $\Big(\dfrac{35}{52}\Big)$

4. $\Big(\dfrac{9}{13}\Big)$

Answer

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

There are 13 spade cards and each suit has 1 king.

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

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

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

Let E be the event of choosing neither a spade nor a king, then

The number of favorable outcomes to the event E = 36

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{36}{52} = \dfrac{9}{13}$

Hence, option 4 is the correct option.

A card is drawn at random from a well-shuffled pack of 52 cards. The probability that the drawn card is neither a king nor a queen, is:

1. $\Big(\dfrac{2}{13}\Big)$

2. $\Big(\dfrac{3}{13}\Big)$

3. $\Big(\dfrac{10}{13}\Big)$

4. $\Big(\dfrac{11}{13}\Big)$

Answer

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

There are 4 kings and 4 queens in a standard deck.

∴ Total number of kings and queens = 4 + 4 = 8.

∴ Number of cards that are neither a king nor a queen = 52 - 8 = 44.

Let E be the event of drawing a card which is neither a king nor a queen, then

The number of favorable outcomes to the event E = 44

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{44}{52} = \dfrac{11}{13}$

Hence, option 4 is the correct option.

What is the probability that a randomly chosen leap year has 52 Sundays?

1. $\Big(\dfrac{1}{7}\Big)$

2. $\Big(\dfrac{2}{7}\Big)$

3. $\Big(\dfrac{5}{7}\Big)$

4. $\Big(\dfrac{6}{7}\Big)$

Answer

In a leap year, there are 366 days.

366 days = 52 weeks + 2 days

The 7 possible pairs for the 2 extra days are:(Monday, Tuesday)(Tuesday, Wednesday)(Wednesday, Thursday)(Thursday, Friday)(Friday, Saturday)(Saturday, Sunday)(Sunday, Monday)

Total number of possible outcomes = 7

Number of outcomes where a Sunday occurs = 2 [(Saturday, Sunday) and (Sunday, Monday)]

Number of outcomes where a Sunday does not occur = 7 - 2 = 5. {i.e,(Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat)}

Let E be the event that a leap year has exactly 52 Sundays,

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{7}$

Hence, option 3 is the correct option.

The probability that a leap year has 53 Sundays is:

1. $\Big(\dfrac{1}{7}\Big)$

2. $\Big(\dfrac{2}{7}\Big)$

3. $\Big(\dfrac{3}{7}\Big)$

4. $\Big(\dfrac{4}{7}\Big)$

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)).

Let E be the event that a leap year has 53 Sundays.

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{7}$

Hence, option 2 is the correct option.

The probability of a non-leap year having 53 Mondays is:

1. $\Big(\dfrac{1}{7}\Big)$

2. $\Big(\dfrac{2}{7}\Big)$

3. $\Big(\dfrac{5}{7}\Big)$

4. $\Big(\dfrac{6}{7}\Big)$

Answer

In a non-leap year (an ordinary year), there are 365 days.

365 days = 52 weeks + 1 day

This 1 extra day can be any of the following 7 possibilities: {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}

Total number of possible outcomes = 7

Number of favourable outcomes (The extra day being a Monday) = 1 (i.e., {Monday}).

Let E be the event that a non-leap year has 53 Mondays,

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{7}$

Hence, option 1 is the correct option.

In a simultaneous throw of two coins, the probability of getting at least one head is:

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{1}{3}\Big)$

3. $\Big(\dfrac{2}{3}\Big)$

4. $\Big(\dfrac{3}{4}\Big)$

Answer

When two coins are thrown simultaneously, the possible outcomes are: {(H, H), (H, T), (T, H), (T, T)}

Total number of outcomes = 4

Let E be the event of getting at least one head,

E = {(H, H), (H, T), (T, H)}

The number of favorable outcomes to the event E = 3

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{4}$

Hence, option 4 is the correct option.

Two fair coins are tossed simultaneously. What is the probability of getting at most one tail?

1. $\Big(\dfrac{1}{4}\Big)$

2. $\Big(\dfrac{1}{2}\Big)$

3. $\Big(\dfrac{3}{4}\Big)$

4. $\Big(\dfrac{3}{8}\Big)$

Answer

When two coins are tossed simultaneously, the possible outcomes are: {(H, H), (H, T), (T, H), (T, T)}

Total number of outcomes = 4

Let E be the event of getting at most one tail,

E = {(H, H), (H, T), (T, H)}

The number of favorable outcomes to the event E = 3

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{4}$

Hence, option 3 is the correct option.

Three unbiased coins are tossed together. What is the probability of getting at least two heads?

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{5}{8}\Big)$

3. $\Big(\dfrac{3}{4}\Big)$

4. $\Big(\dfrac{7}{8}\Big)$

Answer

When three coins are tossed together, the possible outcomes are: {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)}

Total number of outcomes = 8

Let E be the event of getting at least two heads,

E={(H, H, H), (H, H, T), (H, T, H), (T, H, H)}

The number of favorable outcomes to the event E = 4

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{8} = \dfrac{1}{2}$

Hence, option 1 is the correct option.

If three different coins are tossed together, find the probability of getting two heads.

1. $\Big(\dfrac{3}{8}\Big)$

2. $\Big(\dfrac{1}{2}\Big)$

3. $\Big(\dfrac{3}{4}\Big)$

4. $\Big(\dfrac{5}{8}\Big)$

Answer

When three coins are tossed together, the possible outcomes are: {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)}

Total number of outcomes = 8

Let E be the event of getting exactly two heads,

E = {(H, H, T), (H, T, H), (T, H, H)}

The number of favorable outcomes to the event E = 3

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{8}$

Hence, option 1 is the correct option.

Three unbiased coins are tossed together. What is the probability of getting at most two heads?

1. $\Big(\dfrac{1}{4}\Big)$

2. $\Big(\dfrac{3}{4}\Big)$

3. $\Big(\dfrac{3}{8}\Big)$

4. $\Big(\dfrac{7}{8}\Big)$

Answer

When three coins are tossed together, the possible outcomes are: {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)}

Total number of outcomes = 8

Let E be the event of getting at most two heads,

E = {(H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)}

The number of favorable outcomes to the event E = 7

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{8}$

Hence, option 4 is the correct option.

In a simultaneous throw of two dice, what is the probability of getting a total of 7?

1. $\Big(\dfrac{1}{6}\Big)$

2. $\Big(\dfrac{1}{4}\Big)$

3. $\Big(\dfrac{2}{3}\Big)$

4. $\Big(\dfrac{3}{4}\Big)$

Answer

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a total of 7,

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

The number of favorable outcomes to the event E = 6

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}$

Hence, option 1 is the correct option.

Two different dice are rolled together. The probability of getting a sum of 9 is:

1. $\Big(\dfrac{1}{6}\Big)$

2. $\Big(\dfrac{1}{8}\Big)$

3. $\Big(\dfrac{1}{9}\Big)$

4. $\Big(\dfrac{1}{12}\Big)$

Answer

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a sum of 9,

E = {(3, 6), (4, 5), (5, 4), (6, 3)}

The number of favorable outcomes to the event E = 4

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{36} = \dfrac{1}{9}$

Hence, option 3 is the correct option.

In a simultaneous throw of two dice, what is the probability of getting a doublet?

1. $\Big(\dfrac{1}{4}\Big)$

2. $\Big(\dfrac{2}{3}\Big)$

3. $\Big(\dfrac{1}{6}\Big)$

4. $\Big(\dfrac{3}{7}\Big)$

Answer

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a doublet,

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

The number of favorable outcomes to the event E = 6

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}$

Hence, option 3 is the correct option.

In a simultaneous throw of two dice, what is the probability of getting a total of 10 or 11?

1. $\Big(\dfrac{1}{4}\Big)$

2. $\Big(\dfrac{1}{6}\Big)$

3. $\Big(\dfrac{7}{12}\Big)$

4. $\Big(\dfrac{5}{36}\Big)$

Answer

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a total of 10 or 11,

E = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5)}

The number of favorable outcomes to the event E = 5

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{36}$

Hence, option 4 is the correct option.

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{3}{4}\Big)$

3. $\Big(\dfrac{3}{8}\Big)$

4. $\Big(\dfrac{5}{16}\Big)$

Answer

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a product that is even,

The outcomes where both numbers are odd are: (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)

Number of outcomes with an odd product = 9

∴ Number of outcomes with an even product = 36 - 9 = 27.

The number of favorable outcomes to the event E = 27

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{27}{36} = \dfrac{3}{4}$

Hence, option 2 is the correct option.

Two different dice are rolled together. The probability that the product of the numbers appeared is less than 18 is:

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{2}{3}\Big)$

3. $\Big(\dfrac{7}{9}\Big)$

4. $\Big(\dfrac{13}{18}\Big)$

Answer

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

The pairs with a product ≥ 18 = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}

Total number of outcomes where product ≥ 18 = 10

∴ Total number of outcomes where product is less than 18 = 36 - 10 = 26.

Let E be the event that the product is less than 18, then

The number of favorable outcomes to the event E = 26

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{26}{36} = \dfrac{13}{18}$

Hence, option 4 is the correct option.

A bag contains 14 balls out of which x are white. If 6 more white balls are added to the bag, the probability of drawing a white ball is $\Big(\dfrac{1}{2}\Big)$. The value of x is:

1. 4

2. 6

3. 8

4. 10

Answer

Given,

The bag contains 14 balls.

Number of white balls = x

When 6 more white balls are added:

New number of white balls = x + 6

Total number of balls = 14 + 6 = 20

The probability of drawing a white ball now is $\dfrac{1}{2}$.

∴ $P(\text{White ball}) = \dfrac{\text{Number of white balls}}{\text{Total number of balls}} = \dfrac{1}{2} = \dfrac{x + 6}{20}$

2(x + 6) = 20

x + 6 = $\dfrac{20}{2}$

x + 6 = 10

x = 10 - 6

x = 4

Hence, option 1 is the correct option.

The king and queen of diamonds are removed from a pack of well-shuffled cards. One card is selected at random from the remaining cards. The probability of getting the king of clubs is:

1. $\Big(\dfrac{1}{4}\Big)$

2. $\Big(\dfrac{3}{4}\Big)$

3. $\Big(\dfrac{1}{50}\Big)$

4. $\Big(\dfrac{3}{52}\Big)$

Answer

A standard deck of cards contains 52 cards.

In this case, 2 cards (the King of Diamonds and the Queen of Diamonds) are removed from the deck.

Total number of outcomes = 50

Let E be the event of getting the card king of clubs,

The number of favorable outcomes to the event E = 1

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{50}$

Hence, option 3 is the correct option.

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, the number of blue balls in the bag is:

1. 10

2. 15

3. 20

4. 25

Answer

Let the number of blue balls in the bag be x.

Total number of balls in bag = x + 5

The probability of drawing a red ball and a blue ball are:

$P_r = \dfrac{5}{x + 5} \\[1em] P_b = \dfrac{x}{x + 5}$

The probability of drawing a blue ball is thrice (3 times) that of a red ball: P_b = 3 × P_r

Substituting values we get:

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

The number of blue balls in the bag is 15.

Hence, option 2 is the correct option.

The probability of getting a defective pen in a lot of 600 is 0.045. The number of defective pens in the lot is :

1. 27

2. 270

3. 36

4. 360

Answer

The total number of pens in the lot is 600. Let no. of defective pens be x.

The probability of getting a defective pen, P(E) = 0.045.

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \\[1em] \Rightarrow 0.045 = \dfrac{x}{600} \\[1em] \Rightarrow 0.045 = \dfrac{x}{600} \\[1em] \Rightarrow x = 0.045 \times 600 = 27.$

Hence, option 1 is the correct option.

Aditya has a bag containing some black and some white balls. The probability of a randomly chosen ball from this bag being white is $\Big(\dfrac{2}{5}\Big)$. If 5 white balls are added and 5 black balls are removed from the bag, the probability of choosing a white ball changes to $\Big(\dfrac{3}{5}\Big)$. The total number of balls in the bag is:

1. 10

2. 25

3. 45

4. 50

Answer

Let W be the initial number of white balls and B be the initial number of black balls. The total number of balls,

T = W + B

Given,

The probability of picking a white ball is $\dfrac{2}{5}$.

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \\[1em] \Rightarrow \dfrac{W}{T} = \dfrac{2}{5} \\[1em] \Rightarrow 5W = 2T$

Since 5 white balls are added and 5 black balls are removed,

The total number of balls remain same and number of white balls increase W + 5.

Given,

The new probability choosing a white ball is $\dfrac{3}{5}$.

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \\[1em] \Rightarrow \dfrac{W + 5}{T} = \dfrac{3}{5} \\[1em] \Rightarrow 5(W + 5) = 3T \\[1em] \Rightarrow 5W + 25 = 3T \\[1em] \Rightarrow (2T) + 25 = 3T \\[1em] \Rightarrow 25 = 3T - 2T \\[1em] \Rightarrow T = 25.$

Hence, option 2 is the correct option.

Directions Two different dice are rolled together.

Based on this information, answer the following questions:

49. The probability that the sum of the two numbers on the top of the dice is at least 10 is:

(a) $\Big(\dfrac{1}{2}\Big)$

(b) $\Big(\dfrac{1}{3}\Big)$

(c) $\Big(\dfrac{1}{4}\Big)$

(d) $\Big(\dfrac{1}{6}\Big)$

50.The probability of getting two different numbers on the two dice is:

(a) $\Big(\dfrac{1}{6}\Big)$

(b) $\Big(\dfrac{5}{6}\Big)$

(c) $\Big(\dfrac{1}{4}\Big)$

(d) $\Big(\dfrac{3}{4}\Big)$

51.The probability that the product of the two numbers on the top of the dice is at most 12 is:

(a) $\Big(\dfrac{1}{6}\Big)$

(b) $\Big(\dfrac{2}{3}\Big)$

(c) $\Big(\dfrac{23}{36}\Big)$

(d) $\Big(\dfrac{25}{36}\Big)$

52.The probability of getting a multiple of 2 on one die and a multiple of 3 on the other is:

(a) $\Big(\dfrac{1}{3}\Big)$

(b) $\Big(\dfrac{5}{18}\Big)$

(c) $\Big(\dfrac{11}{36}\Big)$

(d) $\Big(\dfrac{13}{36}\Big)$

Answer

49.When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting the the sum is at least 10,

E = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}

The number of favorable outcomes to the event E = 6

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}$

Hence, option (d) is the correct option.

50. When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let A be the event of getting the same number on both dice (doubles),

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

Number of outcomes for A = 6

Let E be the event of getting two different numbers on the two dice,

Number of favorable outcomes to the event E = 36 - 6 = 30

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{30}{36} = \dfrac{5}{6}$

Hence, option (b) is the correct option.

51. When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event that the product of the two numbers is at most 12,

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

The number of favorable outcomes to the event E = 23

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{23}{36}$

Hence, option (c) is the correct option.

52. When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a multiple of 2 on one die and a multiple of 3 on the other,

E = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2), (6, 4)}

The number of favorable outcomes to the event E = 11

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{11}{36}$

Hence, option (c) is the correct option.

Directions A bag contains 5 yellow, 6 red, 3 white and n black balls. The probability of drawing a white ball from the bag is $\Big(\dfrac{1}{7}\Big)$.

Based on this information, answer the following questions:

53.How many black balls are there in the bag?
(a) 6
(b) 7
(c) 8
(d) 9

54.If a ball is picked at random from the bag, what is the probability that the chosen ball is not black?

(a) $\Big(\dfrac{2}{3}\Big)$

(b) $\Big(\dfrac{2}{7}\Big)$

(c) $\Big(\dfrac{4}{7}\Big)$

(d) $\Big(\dfrac{6}{7}\Big)$

55.If a ball is picked at random from the bag, what is the probability that it is either yellow or black?

(a) $\Big(\dfrac{3}{7}\Big)$

(b) $\Big(\dfrac{4}{7}\Big)$

(c) $\Big(\dfrac{6}{7}\Big)$

(d) $\Big(\dfrac{5}{14}\Big)$

56.If 5 more blue balls are added to the bag and one red ball is removed from it, what is the probability of picking up a red ball if a ball is picked up at random from the bag?

(a) $\Big(\dfrac{1}{3}\Big)$

(b) $\Big(\dfrac{1}{5}\Big)$

(c) $\Big(\dfrac{1}{6}\Big)$

(d) $\Big(\dfrac{1}{7}\Big)$

Answer

53. Total number of balls in bag = 5 yellow + 6 red + 3 white + n black = 14 + n

Let E be the event of drawing a white ball.

The number of favorable outcomes (white balls) = 3

$\therefore P(E) = \dfrac{\text{Number of white balls}}{\text{Total number of balls}} \\[1em] \dfrac{1}{7} = \dfrac{3}{14 + n} \\[1em] 14 + n = 21 \\[1em] n = 7.$

Hence, option (b) is the correct option.

54. Total number of balls in bag = 21

Let E be the event that the ball is not black.

Number of favorable outcomes (yellow + red + white) = 5 + 6 + 3 = 14

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{14}{21} = \dfrac{2}{3}$

Hence, option (a) is the correct option.

55. Total number of outcomes = 21

Let E be the event that the ball is either yellow or black.

Number of favorable outcomes (5 yellow + 7 black) = 12

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{21} = \dfrac{4}{7}$

Hence, option (b) is the correct option.

56.Total number of balls in bag = 21

After adding 5 blue balls and removing 1 red ball,

Total number of balls in bag = 25

Number of red balls remaining = 5

Let E be the event of picking a red ball.

The number of favorable outcomes = 5

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$

Hence, option (b) is the correct option.

Directions
Three unbiased coins are tossed simultaneously.

Based on this information, answer the following questions:

57.The probability of getting at least one head is:

(a) $\Big(\dfrac{3}{8}\Big)$

(b) $\Big(\dfrac{1}{2}\Big)$

(c) $\Big(\dfrac{5}{8}\Big)$

(d) $\Big(\dfrac{7}{8}\Big)$

58.The probability of getting at least two tails is:

(a) $\Big(\dfrac{1}{4}\Big)$

(b) $\Big(\dfrac{3}{8}\Big)$

(c) $\Big(\dfrac{1}{2}\Big)$

(d) $\Big(\dfrac{5}{8}\Big)$

59.The probability of getting at most two heads is:

(a) $\Big(\dfrac{1}{2}\Big)$