Probability — Chapter Test

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 consists 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) = $\dfrac{6}{86} = \dfrac{3}{43}.$

Hence, the probability that the number selected is a perfect square number is $\dfrac{3}{43}.$

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is four times that of a red ball, find the number of balls in the bag.

Answer

Let no. of blue balls be x, so total no. of balls = x + 5.

Given,

P(a blue ball drawn) = 4 × P(a red ball drawn) .

$\Rightarrow \dfrac{x}{x + 5} = 4 \times \dfrac{5}{x + 5} \\[1em] \Rightarrow \dfrac{x}{x + 5} = \dfrac{20}{x + 5} \\[1em] \Rightarrow x = 20.$

Total no. of balls = x + 5 = 20 + 5 = 25.

A bag contains 18 balls out of which x balls are white.

(i) If one ball is drawn at random from the bag, what is the probability that it is white ball?

(ii) If 2 more white balls are put in the bag, the probability of drawing a white ball will be $\dfrac{9}{8}$ times that of probability of white ball coming in part (i). Find the value of x.

Answer

(i) Let A be the event 'a white ball is drawn'.

∴ The no. of favourable outcomes to event A = x.

∴ P(a white ball is drawn) = $\dfrac{x}{18}.$

Hence, the probability that a white ball is drawn = $\dfrac{x}{18}$.

(ii) If 2 more white balls are added,

No. of white balls = x + 2,

Total no. of balls = 18 + 2 = 20.

Hence now,

P(a white ball is drawn) = $\dfrac{x + 2}{20}$

Given, probability of drawing a white ball will be $\dfrac{9}{8}$ times that of probability of white ball coming in part (i)

$\therefore \dfrac{x + 2}{20} = \dfrac{9}{8} \times \dfrac{x}{18} \\[1em] \Rightarrow \dfrac{x + 2}{20} = \dfrac{x}{16} \\[1em] \Rightarrow 16(x + 2) = 20x \\[1em] \Rightarrow 16x + 32 = 20x \\[1em] \Rightarrow 20x - 16x = 32 \\[1em] \Rightarrow 4x = 32 \\[1em] \Rightarrow x = 8.$

Hence, the value of x = 8.

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

(i) a red face card

(ii) neither a club nor a spade

(iii) neither an ace nor a king of red colour

(iv) neither a red card nor a queen

(v) neither a red card nor a black king.

Answer

Well shuffling ensures equally likely outcomes.

Total number of outcomes = 52.

(i) Each suit has one king, one queen and one jack, and there are two suits of red colour.

∴ There are 2 kings, 2 queens and 2 jacks.

∴ The number of red face cards = 6.

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

Hence, the probability that a card drawn is a red face card = $\dfrac{3}{26}$.

(ii) There are 13 clubs and 13 spades i.e. total = 26 cards.

No. of cards left other than spades and clubs = 52 - 26 = 26.

P(neither a club nor spade) = $\dfrac{26}{52} = \dfrac{1}{2}.$

Hence, the probability that a card drawn is neither a club nor spade card = $\dfrac{1}{2}$.

(iii) There are 2 kings of red colour, one of heart and one of diamond.

There are 4 aces, one of each suit.

No. of cards other than ace and king of red colour = 52 - 4 - 2 = 46.

P(neither an ace nor a king of red colour) = $\dfrac{46}{52} = \dfrac{23}{26}.$

Hence, the probability that a card drawn is neither an ace nor a king of red colour = $\dfrac{23}{26}$.

(iv) There are 26 red cards, 13 of hearts and 13 of diamonds.

There are 4 queens, one of each suit but since red queens are included in red cards hence, queens left = 2.

Total no. of red cards and queen = 26 + 2 = 28.

No. of cards other than queen and red cards = 52 - 28 = 24.

P(neither a red card nor queen) = $\dfrac{24}{52} = \dfrac{6}{13}.$

Hence, the probability that a card drawn is neither a red card nor queen = $\dfrac{6}{13}$

(v) There are 26 red cards, 13 of hearts and 13 of diamonds.

There are 2 black kings, one of club and one of spade.

Total no. of red cards and black kings = 26 + 2 = 28.

No. of cards other than red cards and black kings = 52 - 28 = 24.

P(neither a red card nor black king) = $\dfrac{24}{52} = \dfrac{6}{13}.$

Hence, the probability that a card drawn is neither a red card nor black king = $\dfrac{6}{13}$

From pack of 52 playing cards, black jacks, black kings and black aces are removed and then the remaining pack is well-shuffled. A card is drawn at random from the remaining pack. Find the probability of getting

(i) a red card

(ii) a face card

(iii) a diamond or a club

(iv) a queen or a spade.

Answer

There are 2 black jacks, 2 black kings and 2 black aces.

No. of cards left after removing black jacks, black kings and black aces = 52 - 6 = 46.

(i) There are 26 red cards, 13 of hearts and 13 of diamonds.

∴ P(a red card) = $\dfrac{26}{46} = \dfrac{13}{23}.$

Hence, the probability of drawing a red card = $\dfrac{13}{23}.$

(ii) There are 12 face cards, out of which 4 are removed.

No. of face cards left = 8,

P(a face card) = $\dfrac{8}{46} = \dfrac{4}{23}.$

Hence, the probability of drawing a face card = $\dfrac{4}{23}.$

(iii) There are 13 cards in diamond and 13 cards in club.

Out of club, one king, one queen and one ace is removed,

No. of club cards left = 13 - 3 = 10.

Total no. of diamond and club cards = 13 + 10 = 23.

P(a diamond or club) = $\dfrac{23}{46} = \dfrac{1}{2}.$

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

(iv) There are 13 cards in spade and 4 queens (including the queen of spade).

Out of spade, one king, one queen and one ace is removed,

No. of spade cards left = 13 - 3 = 10.

3 queens are left after removing the queen of spade, as it is already included while counting spade cards.

Total no. of spade and queens = 10 + 3 = 13.

P(a spade or queen) = $\dfrac{13}{46}.$

Hence, the probability of drawing a spade or queen = $\dfrac{13}{46}.$

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

(i) sum 7

(ii) sum ≤ 3

(iii) sum ≤ 10

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 sum of 7,

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

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

∴ P(sum of 7) = $\dfrac{6}{36} = \dfrac{1}{6}.$

Hence, the probability of getting a sum of 7 is $\dfrac{1}{6}$.

(ii) Let B be the event of getting a sum ≤ 3,

B = {(1, 1), (1, 2), (2, 1)}

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

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

Hence, the probability of getting a sum ≤ 3 is $\dfrac{1}{12}$.

(iii) Let C be the event of getting a sum ≤ 10,

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

∴ The number of outcomes favourable to event C = 33.

∴ P(sum ≤ 10) = $\dfrac{33}{36} = \dfrac{11}{12}.$

Hence, the probability of getting a sum ≤ 10 is $\dfrac{11}{12}$.

Two dice are thrown together. Find the probability that the product of the numbers on the top of two dice is

(i) 4

(ii) 12

(iii) 7

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 numbers on top of dice with product 4.

A = {(1, 4), (2, 2), (4, 1)}.

∴ P(A) = $\dfrac{3}{36} = \dfrac{1}{12}.$

Hence, the probability that the product of the numbers on the top of two dice is 4 is $\dfrac{1}{12}.$

(ii) Let B be the event of getting numbers on top of dice with product 12.

B = {(2, 6), (3, 4), (4, 3), (6, 2)}.

∴ P(B) = $\dfrac{4}{36} = \dfrac{1}{9}.$

Hence, the probability that the product of the numbers on the top of two dice is 12 is $\dfrac{1}{9}.$

(iii) Let C be the event of getting numbers on top of dice with product 7.

C = {}.

∴ P(C) = $\dfrac{0}{36} = 0.$

Hence, the probability that the product of the numbers on the top of two dice is 7 is 0.