Gujarat Board Statistics Class 12 GSEB Solutions Part 2 Chapter 1 Probability Ex 1.3 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 1 Probability Ex 1.3

Question 1.

2 cards are drawn from a pack of 52 cards. Find the probability that both the cards drawn are

(1) of the same suit,

(2) of the same colour.

Answer:

Total number of primary outcomes of drawing 2 cards from a pack of 52 cards is

n = ^{52}C_{2} = \(\frac{52 \times 51}{2 \times 1}\) = 1326

(1) A = Event that the two cards are of the same suit, i.e., two spade cards or two club cards or two heart cards or two diamond cards.

In a pack of 52 cards each suit has 13 cards.

∴ Favourable outcomes for the event A is

m = ^{13}C_{2} + ^{13}C_{2} + ^{13}C_{2} + ^{13}C_{2}

= 4[^{13}C_{2}] = 4\(\left[\frac{13 \times 12}{2 \times 1}\right]\)

= 4 (78)

= 312

Hence, P(A) = \(\frac{m}{n}\) = \(\frac{312}{1326}\) = \(\frac{4}{17}\)

(2) B = Event that two cards are of the same colour, i.e., two black cards or two red cards.

In a pack of 52 cards, there are 26 black cards and 26 red cards.

∴ Favourable outcomes for the event B is

m = ^{26}C_{2} + ^{26}C_{2} = 2 [^{26}C_{2}]

= \(\left[\frac{26 \times 25}{2 \times 1}\right]\)

= 2 [325]

= 650

Hence, P(B) = \(\frac{m}{n}\) = \(\frac{650}{1326}\) = \(\frac{25}{51}\)

Question 2.

3 books of Statistics and 4 of Mathematics are arranged on a shelf. Two books are randomly selected from these books. Find the probability that both the books selected are of the same subject.

Answer:

On a self there are 3 books of Statistics + 4 books of Mathematics = 7 books.

Total number of primary outcomes of selecting 2 books randomly from 7 books is,

n = ^{7}C_{2} = \(\frac{7 \times 6}{2 \times 1}\) = 21.

A = Event that selected two books are of the same subject, i.e., 2 books of Statistics or 2 books of Mathematics.

∴ Favourable outcomes for the event A is

m = ^{3}C_{2} + ^{4}C_{2}

= \(\frac{3 \times 2}{2 \times 1}\) + \(\frac{4 \times 3}{2 \times 1}\)

= 3 + 6

= 9

Hence, P(A) = \(\frac{m}{n}\) = \(\frac{9}{21}\) = \(\frac{3}{7}\)

Question 3.

One card is randomly drawn from a pack of 52 cards. Find the probability that it is

(1) spade card or ace

(2) neighter spade nor ace.

Answer:

Total number of primary outcomes or drawing one card randomly from a pack of 52 cards is n = ^{52}C_{1} = 52

(1) A = Event that the card drawn is spade card

∴ m = ^{13}C_{1} = 13

∴ P(A) = \(\frac{m}{n}\) = \(\frac{13}{52}\)

B = Event that the card drawn is ace

∴ m = ^{4}C_{1} = 4

∴ P(B) = \(\frac{4}{52}\)

A ∩ B = Event that the card is spade ace.

∴ m = 1

∴ P[B) = \(\frac{m}{n}\) = \(\frac{1}{52}\)

Now, A ∪ B = Event that the card is spade card or ace.

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{13}{52}+\frac{4}{52}-\frac{1}{52}\)

= \(\frac{16}{52}\) = \(\frac{4}{13}\)

(2) A’ = Event that the card is not spade card

B’ = Event that the card is not ace card.

Now, A’ ∩ B’ = Event that the card is neither spade nor ace.

∴ P(A’ ∩ B’) = P(A ∪ B)’

= 1 – P(A ∪ B)

= 1 – \(\frac{4}{13}\)

= \(\frac{9}{13}\)

Question 4.

A number is selected from the natural number 1 to 100. Find the probability of the event that the selected number is a multiple of 3 or 5.

Answer:

Here, U = {1, 2, 3, ………… 100}

A number is selected from U.

∴ n = ^{100}C_{1} = 100

A = Event that the number selected is multiple of 3

= {3, 6, 9, 12, …, 96, 99}

∴ Favourable outcomes for the event A is m = 33.

∴ p(A) = \(\frac{m}{n}\) = \(\frac{33}{100}\)

B = Event that the number selected is multiple of 5

= (5, 10, 15, 20, …, 95, 100}

∴ Favourable outcomes for the event B is m = 20.

∴ p(B) = \(\frac{m}{n}\) = \(\frac{20}{100}\)

A ∩ B = Event that the number selected is multiple of 3 and 5, i.e., multiple of 15

= {15, 30,…, 75, 90}

∴ Favourable outcomes for the event A ∩ B is

m = 6.

∴ P(A ∩ B) = \(\frac{m}{n}\) = \(\frac{6}{100}\)

Now, (A ∪ B) = Event that the number selected is multiple of 3 or 5.

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{33}{100}+\frac{20}{100}-\frac{6}{100}\)

= \(\frac{53-6}{100}\) = \(\frac{47}{100}\)

Question 5.

Two balanced dice are thrown simultaneously. Find the probability that the sum of numbers on two dice is a multiple of 2 or 3.

Answer:

Two balanced dice are thrown simultaneously.

∴ n = 62 = 36

A = Event that the sum of numbers on two dice is a multiple of 2 i.e., 2, 4, 6, 8, 10 or 12.

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

∴ Favourable outcomes for the event A is

m = 18.

∴ P(A) = \(\frac{m}{n}\) = \(\frac{18}{36}\)

B = Event that the sum of numbers on the dice is a multiple of 3 i.e., 3, 6, 9 or 12.

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

∴ Favourable outcomes for the event B is

m = 12.

∴ p(B) = \(\frac{m}{n}\) = \(\frac{12}{36}\)

A ∩ B = Event that the sum of the numbers on dice is a multiple of 2 and 3, i.e., 6 or 12.

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

∴ Favourable outcomes for the event A ∩ B is m = 6.

∴ P(A ∩ B) = \(\frac{m}{n}\) = \(\frac{6}{36}\)

Now, A ∪ B = Event that the sum of numbers on dice is a multiple of 2 or 3.

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{18}{36}+\frac{12}{36}-\frac{6}{36}\)

= \(\frac{24}{36}\) = \(\frac{2}{3}\)

Question 6.

The probability that the price of potato rises in the vegetable market during festive days in 0.8. The probability that the price of onion rises is 0.7. The probability of rise in price of both potato and onion is 0.6. Find the probability of rise in price of at least one of the two, potato and onion.

Answer:

A = Event that the price of potato rises.

B = Event that the price of onion rises.

A ∩ B = Event that the prices of potato and onion rise.

Now, P(A) = 0.8, P(B) = 0.7 and P(A ∩ B) = 0.6 are given.

A ∪ B = Event that the rise in price of at least one of the two, potato and onion.

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.8 + 0.7 – 0.6 = 0.9

Question 7.

Two aircrafts drop bomb to destroy a bridge. The probability that a bomb dropped from the first aircraft hits the target is 0.9 and the probability that a bomb from the second aircraft hits the target is 0.7. The probability of one bomb dropped from both the aircrafts hitting the target is 0.63. The bridge is destroyed even if one bomb drops on it. Find the probability that the bridge is destroyed.

Answer:

A = Event that a bomb dropped from the first aircraft hits the target

B = Event that a bomb dropped from the second aircraft hits the target

A ∩ B = Event that a bomb dropped from both the aircraft hits the target

Now, P(A) = 0.9, P(B) = 0.7, P(A ∩ B) = 0.63 are given.

If one or more bombs dropped on the bridge, it is destroyed, i.e., at least one bomb dropped on the bridge it is destroyed.

∴ A ∪ B = Event that the bridge is destroyed.

Now, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

= 0.9 + 0.7 – 0.63

= 1.6 – 0.63

= 0.97

Question 8.

The, probability that a teenager coming to a restaurant for dinner orders pizza is 0.63. The probability of ordering cold drink is 0.54. The probability that the teenager orders at least one out of pizza and cold drink is 0.88. Find the probability that the teenager coming for dinner on a certain day orders only one of the two items from pizza and cold drink.

Answer:

A = Event that a teenager orders pizza

B = Event that a teenager orders cold drink

A ∪ B = Even that a teenager order at least one of the two, pizza or cold drink.

Now, P(A) = 0.63, P(B) = 0.54 and

P(A ∪ B) = 0.88 are given.

A ∩ B = Event that a teenager orders both pizza and cold drink

Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

P(A ∩ B) = P(A) + P(B)-P(A ∪ B)

= 0.63 + 0.54 – 0.88 = 0.29

C = Event that a teenager orders only one from pizza and cold drink

A ∩ B’ = Event that a teenager orders only for pizza (Event A) and not for cold drink (Event B’)

OR

A’ ∩ B = Event that a teenager orders only for cold drink (Event B) and not for pizza (Event A’)

∴ C = (A ∩ B’) ∪ (A’ ∩ B) ,

(A ∩ B’) and (A’ ∩ B) are mutually exclusive.

∴ P(C) = P(A ∩ B’) + P(A’ ∩ B)

= [P (A) – P(A ∩ B)] + [P(B) – P(A ∩ B)l

= [0.63 – 0.29] + [0.54 – 0.29]

= 0.34 + 0.25 = 0.59

Question 9.

If A and B are mutually exclusive and exhaustive events in a sample space U and P(A) = 2P(B), then find P(A).

Answer:

A and B are exhaustive and mutually exclusive events.

∴ A ∪ B = U

∴ P(A ∪ B) = P(U)

∴ P(A)+ P(B) = 1 (1)

Now, P(A) = 2P(B)

∴ Putting P(B) = \(\frac{P(A)}{2}\) in result (1).

P(A) + \(\frac{P(A)}{2}\) = 1

∴ \(\frac{2 P(A)+P(A)}{2}\) = 1

∴ 3P (A) = 2

∴ P(A) = \(\frac{2}{3}\)

Question 10.

Three events A, B and C in a sample space are mutually exclusive and exhaustive. If 4P (A) = 5P (B) = 3P (C), then find P (A ∪ C) and P(B ∪ C).

Answer:

Take 4P(A) = 5P(B) = 3P(C) = x.

Now, A, B and C are mutually exclusive and exhaustive events,

∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = 1

∴ \(\frac{x}{4}+\frac{x}{5}+\frac{x}{3}\) = 1

∴ \(\frac{15 x+12 x+20 x}{60}\) = 1

∴ 47x = 60

∴ x = \(\frac{60}{47}\)

Putting x = \(\frac{60}{47}\),

P(A) = \(\frac{x}{4}\) = \(\frac{60}{4 \times 47}\) = \(\frac{15}{47}\)

P(B) = \(\frac{x}{5}\) = \(\frac{60}{5 \times 47}\) = \(\frac{12}{47}\)

P(C) = \(\frac{x}{3}\) = \(\frac{60}{3 \times 47}\) = \(\frac{20}{47}\)

P(A ∪ C) = P(A) + P(C)

= \(\frac{15}{47}\) + \(\frac{20}{47}\)

= \(\frac{35}{47}\)

P(B ∪ C) = P(B) + P(C)

= \(\frac{12}{47}\) + \(\frac{20}{47}\)

= \(\frac{32}{47}\)

Question 11.

Find P(A ∪ B ∪ C) using the following information about three events A, B and C in a sample space:

P(A) = 0.65,

P(B) = 0.45,

P(C) = 0.25,

P(A ∩ B) = 0.25,

P(A ∩ C) = 0.15,

P(B ∩ C) = 0.2,

P(A ∩ B ∩ C) = 0.05

Answer:

Here, P(A) = 0.65,

P(B) = 0.45,

P(C) = 0.25,

P(A ∩ B) = 0.25,

P(A ∩ C) = 0.15,

P(B ∩ C) = 0.2,

P(A ∩ B ∩ C) = 0.05 are given.

Now, P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)

= 0.65 + 0.45 + 0.25 – 0.25 – 0.15 – 0.2 + 0.05 = 0.8

Question 12.

Three events A, B and C in a sample space are mutually exclusive and exhaustive. If P(C) = 0.8 and 3P(B) = 2P(A’), then find P(A) and P(B).

Answer:

Here, P(C’) = 0.8, 3P(B) = 2P(A’) are given.

P(C’) = 0.8 ∴ P(C) = 1 – P(C’) = 1 – 0.8 = 0.2

3P(B) = 2P(A)

∴ 3P(B) = 2[1 – P(A)]

∴ 3P(B) = 2 – 2 P(A)

∴ p(B) = \(\frac{2-2 P(A)}{3}\)

A, B and C are mutually exclusive and exhaustive events.

∴ P(A ∪ B ∪ C) = 1 .

∴ P(A) + P(B) + P(C) = 1

∴ P(A) + \(\frac{2-2 P(A)}{3}\) + 0.2 = 1

∴ \(\frac{3 P(A)+2-2 P(A)}{3}\) = 1 – 0.2

∴ \(\frac{P(A)+2}{3}\) = 0.8

∴ P(A) + 2 = 0.8 × 3 = 2.4

∴ P(A) = 2.4 – 2 = 0.4

Now, putting P (A) = 0.4 in P (B) = \(\frac{2-2 P(A)}{3}\),

P (B) = \(\frac{2-2(0.4)}{3}\) = \(\frac{2-0.8}{3}\) = \(\frac{1.2}{3}\) = 0.4