Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 13 Probability Ex 13.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 13 Probability Ex 13.1

Question 1.

Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E/F) and P(F/E).

Solution:

(i) P(E/F) = \(\frac{E∩F}{P(F)}\) = \(\frac{0.2}{0.3}\) = \(\frac{2}{3}\).

(ii) P(F/E) = \(\frac{P(E∩F)}{P(E)}\) = \(\frac{0.2}{0.6}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\).

Question 2.

Compute P(A/B), if P(B) = 0.5 and P(A ∩ B) = 0.32.

Solution:

P(A/B) = \(\frac{P(A∩B)}{P(B)}\) = \(\frac{0.32}{0.5}\) = \(\frac{32}{50}\) = \(\frac{16}{25}\).

Question 3.

If P(A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4, find

(i) P(A ∩ B)

(ii) P(A/B)

(iii) P(A ∪ B)

Solution:

(i) P(B/A) = \(\frac{P(A∩B)}{P(B)}\) ⇒ 0.4 = \(\frac{P(A∩B)}{0.8}\)

∴ P(A ∩ B) = 0.4 × 0.8 = 0.32.

(ii) P(A/B) = \(\frac{P(A∩B)}{P(B)}\) = \(\frac{0.32}{0.5}\) = \(\frac{32}{50}\) = \(\frac{16}{25}\).

(iii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.8 + 0.5 – 0.32 = 1.30 – 0.32

= 0.98.

Question 4.

Evaluate P(A ∪ B), if 2P(A) = P(B) = \(\frac{5}{13}\) and P(A/B) = \(\frac{2}{5}\).

Solution:

We have:

2P(A) = P(B) = \(\frac{5}{13}\)

⇒ P(A) = \(\frac{5}{26}\), P(B) = \(\frac{5}{13}\).

So, P(A/B) = \(\frac{P(A∩B)}{P(B)}\)

or \(\frac{2}{5}\) = \(\frac{P(A \cap B)}{\frac{5}{13}}\)

⇒ P(A ∩ B) = \(\frac{2}{5}\) × \(\frac{5}{13}\) = \(\frac{2}{13}\).

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{5}{26}\) + \(\frac{5}{13}\) – \(\frac{2}{13}\) = \(\frac{5+10-4}{26}\) = \(\frac{11}{26}\).

Question 5.

If P(A) = \(\frac{6}{11}\), P(B) = \(\frac{5}{11}\) and P(A ∪B) = \(\frac{7}{11}\), find

(i) P(A ∩ B)

(ii) P(A/B)

(iii) P(B/A)

Solution:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

(i) P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

= \(\frac{6}{11}\) + \(\frac{5}{11}\) – \(\frac{7}{11}\) = \(\frac{4}{11}\).

(ii)

(iii)

Determine P(E/F) in the questions 6 to 9:

Question 6.

A coin is tossed three times, where

(i) E : head on third toss , F : Heads on first two tosses

(ii) E : at least two heads, F : at most two heads

(iii) E : at most two tails F : at least one tail.

Solution:

(i) E = Head occurs on third toss as

{HHH, HTH, THH, TTH}

F : Heads on first two tosses = {HHH, HHT}

∴ E ∩ F = {HHH}

(ii) E : At least two heads = {HHT, HTH, THH, HHH}

F : At most two heads = {TTT, HTT, THT, HTT, HHT, HTH, THH}

∴ E ∩ F : {HHT, HTH, THH}

(iii) E : At most two tails = {HHT, THT, TTH, HHT, HTH, THH, HHH}

F : At least one tail = {THH, HTH, HHT, TTH, THT, HTT, TTT}

∴ E ∩ F : {HTT, THT, TTH, THH, HTH, HHT}

Question 7.

Two coins are tossed once, where

(i) E : tail appear on one coin ,F : one coin shows head

(ii) E : no tail appears, F : no head appears.

Solution:

(i) E : Tail appears on one coin = {TH, HT}

F : One coin shows head = {HT, TH}

∴ E ∩ F : {TH, HT}

(ii) E : no tail appears = {HH}

F : no head appears = {TT}

So, E ∩ F = ∅

∴ P(E/F) = \(\frac{P(E∩F)}{P(F)}\) = 0.

Question 8.

A dice is thrown three times, where

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.

Solution:

A dice is thrown three times.

E : 4 appears on third less

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

(2, 1, 4, (2, 2, 4), (2, 3, 4) ……………… (2, 6, 4)

(3, 1, 4), (3, 2, 4), (3, 3, 4) …………………. (3, 6, 4)

(4, 1, 4), (4, 2, 4), (4, 3, 4) ………………….. (4, 6, 4)

(5, 1, 4), (5, 2, 4), (5, 3, 4) …………………. (5, 6, 4)

(6, 1, 4), (6, 2, 4), (6, 3, 4) …………………… (6, 6, 4)}

These are 36 cases.

F : 6 and 5 appears respectively on first two tosses

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

These are six cases.

Here, E ∩ F = {6, 5, 4}

Question 9.

Mother, father and son line up at random for a family picture, such that:

E : Son on one end

F : Father in middle

Solution:

Mother (m), Father (f) and Son (s) line up at random.

E : Son on one end : {(s, m, f), (s, f, m), (f, m, s), (m, f, s)}

F : Father in middle : {(m, f, s), (s, f, m)}

Question 10.

A black and red dice are rolled:

(a) Find the conditional probability of obtaining a sum greater than 9, given that the black dice resulted in a 5.

(b) Find the conditional probability of obtaining the sum of 8, given that the red dice resulted in a number less than 4.

Solution:

(a) When two dice are thrown, a sum greater than 9 may be obtained as

A : {4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}

B : Black dice will result in 5.

⇒ 5 is on black dice, any other number on red dice, i.e.,

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

So, A ∩ B = {(5, 5), (5, 6)}

(b) A : Sum 8 is obtained = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

B : red dice results in a number less than 4.

Either first or second dice is red

B = {(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)}

So, A ∩ B = {(2, 6), (3, 5)}

Question 11.

A fair dice is rolled. Consider the event E = {1, 3, 5} F = {2, 3} and G = {2, 3, 4, 5}. Find

(i) P(E/F) and P(F/E)

(ii) P(E/G), P(G/E)

(iii) P(E ∪ F/G) and P(E ∩ F/G)

Solution:

(i) E = {1, 3, 5}, F = {2, 3}, E ∩ F = {3}.

(ii) E = {1, 3, 5}, G = {2, 3, 4, 5}, E ∩ G = {3, 5}.

(iii) E = {1, 3, 5}, F = {2, 3}, G = {2, 3, 4, 5}

E ∩ G = {3, 5}, F ∩ G = {2, 3}, (E ∩ F) ∩ G = {3}

∴ P(E ∩ G) = \(\frac{2}{6}\), P(F ∩ G) = \(\frac{2}{6}\), P[(E ∩ F) ∩ G] = \(\frac{1}{6}\)

Now, P(E ∪ F/G) = P(E/G) + P(F/G) – P[(E ∩ F)/G]

Question 12.

Assume that each child bom is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls, given that

(i) the youngest is a girl?

(ii) at least one is a girl?

Solution:

First and second girls are denoted by G_{1}, G_{2} and boys by B_{1}, B_{2}

Sample space S = {(G_{1}G_{2}), (G_{1}B_{2}), (B_{1}G_{2}), (B_{1}B_{2})}

Let A = Both the children are girls = {G_{1}G_{2}}

B = Youngest child is a girl = {G_{1}G_{2}, B_{1}G_{2}}

C = At least one is a girl = {G_{1}B_{2}, G_{1}G_{2}, B_{1}G_{2}}

So, A ∩ B = {G_{1}G_{2}}, A ∩ C = {G_{1}G_{2}}

Question 13.

An instructor has a test bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions.

If a question is selected at random from the test bank, what is the probability that it will be an easy questions given that it is a multiple choice questions?

Solution:

The given data may be tabulated as follows:

Let us denote E = Easy questions, D = Difficult questions, T = True/False questions and M = Multiple choice questions.

Number of easy multiple choice questions = 500.

Total number of questions = 1400.

P(E ∩ M) = Probability of selecting an easy and multiple choice question

= \(\frac{500}{1400}\).

Total number of multiple choice questions

= 500 + 400 = 900

P(M) = Probability of selecting one multiple choice question

= \(\frac{900}{1400}\)

∴ P(E/M) = \(\frac{P(E∩M)}{P(M)}\) = \(\frac{500}{1400}\) ÷ \(\frac{900}{1400}\) = \(\frac{5}{9}\).

Question 14.

Given that two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of the numbers on the dice is 4’.

Solution:

When the numbers appearing on throwing two dice are different.

⇒ The numbers are not the same

Total number of exhaustive cases = 36

Number of cases when doublets do not occur = 36 – 6 = 30

Cases when the sum is 4 are {(1, 3), (2, 2), (3, 1)}

Let A denotes the event when the sum of numbers on two dice is 4 and B is the event when the numbers appearing on the dice are different.

∴ A ∩ B = {(1, 3), (3, 1)}

So, P(A ∩ B) = \(\frac{2}{36}\), P(B) = \(\frac{30}{36}\).

∴ P(A/B) = \(\frac{P(A∩B)}{P(B)}\) = \(\frac{2}{36}\) ÷ \(\frac{3}{36}\) = \(\frac{1}{15}\).

Question 15.

Consider the experiment of throwing a dice, ‘if a multiple of 3 comes up, throw the dice again and if any other number comes toss a coin’.

Find the conditional probability of the event ‘the coin shows tail’, given that at least one dice shows a 3.

Solution:

Let there be n throws in which a multiple of 3 occurs every time.

Probability of getting a multiple of 3 (i.e., 3 or 6) in one throw

= \(\frac{2}{6}\) = \(\frac{1}{3}\).

∴ Probability of getting a multiple of 3 in n throws = (\(\frac{1}{3}\))^{n}

Probability of getting a 6 in one throw = \(\frac{1}{6}\)

∴ Probability of getting a 6 in n throws = (\(\frac{1}{6}\))^{n}.

⇒ Probability of getting at least a 3 in n throws = (\(\frac{1}{3}\))^{n} – (\(\frac{1}{6}\))^{n}

Let a multiple of 3 does not occur in (n + 1)th throw.

∴ Probability of getting 1, 2, 4, 5 (not a multiple of 3) in

(n + 1)^{th} throw = \(\frac{4}{6}\) = \(\frac{2}{3}\).

In the next throw, a coin is tossed and tail occurs.

∴ Probability of getting a tail = \(\frac{1}{2}\)

⇒ Probability of getting at least a 3 and a tail in the end in

Question 16.

If P(A) = \(\frac{1}{2}\), P(B) = 0, then P(A/B) is

(A) 0

(B) \(\frac{1}{2}\)

(C) not defined

(D) 1

Solution:

P(A) = \(\frac{1}{2}\), P(B) = 0

P(A/B) = \(\frac{P(A∩B)}{P(B)}\) = \(\frac{P(A∩B)}{0}\) = Not defined.

∴ Part (C) is the correct answer.

Question 17.

If A and B are events such that P(A/B) = P(B/A), then

(A) A ⊂ B

(B) B = A

(C) A ∩ B = ∅

(D) P(A) = P(B)

Solution:

P(A/B) = P(B/A)

⇒ \(\frac{P(A∩B)}{P(B)}\) = \(\frac{P(A∩B)}{P(A)}\)

⇒ P(A) = P(B)

∴ Part (D) is the correct answer.