Class 12 NCERT Solutions- Mathematics Part ii – Chapter 13 – Probability Miscellaneous Exercise on Chapter 13

Question 1. A and B are two events such that P (A) ≠ 0. Find P (B|A), if:

(i) A is a subset of B ​

(ii) A ∩ B = φ

Solution:

It is given that,

A and B are two events such that P (A) ≠ 0

We have, A ∩ B = A

P( A ∩ B) = P(B ∩ A) = P(A)

Hence , P(B | A)= P(B ∩ A) /P(A)

P(A) /P(A)

= 1

(ii) We have,

P (A ∩ B) = 0

P(B | A) = P(A ∩ B) /P(A)

= 0

Question 2. A couple has two children,

(i) Find the probability that both children are males, if it is known that at least one of the children is male.

(ii) Find the probability that both children are females, if it is known that the elder child is a female.

Solution:

(i) According to the question, if the couple has two children, then the sample space is

S = {(b, b), (b, g), (g, b), (g, g)}

Assume that A denote the event of both children having male and B denote the event of having at least one of the male children.

Thus, we have

A ∩ B = {(b, b)}

P (A ∩ B ) = 1/4

P(A) = 1/4

P(B) = 3/4

Hence , P(A/B) = P (A ∩ B )/P (B)

By substituting the values we get

=1/4/3/4

1/3

(ii) Assume that C denotes the event of having both children as females and D denotes the event of having an elder child as female.

∴ C = {(g, g)}

P (C) = ¼

And, D = {(g, b), (g, g)}

P (D) = (2/4)

Hence, P (C/D) = P ( C ∩D ) / P (D)

= 1/4/2/4

=1/2

Question 3. Suppose that 5% of men and 0.25% of women have grey hair. A grey-haired person is selected at random. What is the probability of this person being male? Assume that there is an equal number of males and females.

Solution:

Given that 5% of men and 0.25% of women have grey hair.

Total % of people having grey hair = 5 + 0.25

= 5.25 %

Hence, the probability of having a selected male person having grey hair, P = 5/25 = 20/21

Question 4 . Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Solution:

Given that 90% of the people are right-handed.

Let p denotes the probability of people that are right-handed, and q denotes the probability of people that are left handed.

p = 9/10 and q = 1 – 9/10 = 1/10

Now, by using the binomial distribution probability of having more than 6 right-handed people can be given as:

∑_r=7^{10 10}C_r P ^r q ^n-r= ∑_r=7^{10 10}C_r (9/10)^r (1/10)^10-r

Hence, the Probability of having more than 6 right handed people:

= 1 – P (More than 6 people are right handed)

= 1 – ∑_r=7^{10 10}C_r (0.9)^r (0.1)^10-r

Question 5. If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?

Solution:

We know that in a leap year, there are a total of 366 days, 52 weeks and 2 days.

Now, in 52 weeks, there is a total of 52 Tuesdays.

∴ The probability that the leap year will contain 53 Tuesdays is equal to the probability of remaining 2 days will be

Tuesdays.

Thus, the remaining two days can be

(Monday and Tuesday), (Tuesday and Wednesday), (Wednesday and Thursday), (Thursday and Friday), (Friday and

Saturday), (Saturday and Sunday) and (Sunday and Monday)

∴ Total Number of cases = 7

Cases in which Tuesday can come = 2

Hence, probability (leap year having 53 Tuesdays) = 2/7

Question 6. Suppose we have four boxes A, B, C, D containing colored marble as given below:

BOX	RED	WHITE	BLACK
A	1	6	3
B	6	2	2
C	8	1	1
D	0	6	4

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

Solution:

Let us assume R the event of drawing the red marbles.

Let us also assume EA ,EB and EC denote the boxes A , B and C , respectively .

Given that ,

Total number of marbles = 40

Also , total number of red marbles = 15

P (R) = 15/40

3/8

Probability of taking out the red marble from box A ,

P( E_A/R) = P (E_B∩ R)/ P(R)

= 6/40/3/8

= 2/5

And , Probability of taking out the red marble from box c,

P( E_c/R) = P (E _c∩ R)/ P(R)

= 8/40/3/8

=8/15

Question 7. Assume that the chances of a patient having a heart attack are 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30%, and a prescription of a certain drug reduces its chances by 25%. At a time, a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.

Solution :

Let us assume X denotes the events having a person ‘s heart attack.

A , denote events having the selected person follow the course of yoga and meditation .

And A ,denote the events having the person adopted the drug prescription .

It is given in the question that ,

P(X) = 0.40

P (A₁) = P (A₂) = 1/2

And P(X|A₁ ) = 0.40 X 0.70 = 0.28

P (X|A₂) = 0.40 X 0.75 = 0.30

Probability (The patient suffering from a heart attack and followed a course of meditation and ⁄yoga ) :

p (A₁ | X ) = P(A₁) P (X|A₁)\ P(A₁) P ( X|A₁) + P (A₂) P(X|A₂)

= 1/2 X 0.28/1/2 X 0.28 + 1/2 X 0.30

= 14/29

Question 8. If each element of a second-order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 1/2).

Solution :

From the question , we have

Total number of determinants of second order where the element being or 1 = (2)⁴ = 16

Now , we have the value of determinants is positive in the following cases .

[Tex]\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix}[/Tex]

Question 9 . An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P (A fails) = 0.2

P (B fails alone) = 0.15

P (A and B fail) = 0.15

Evaluate the following probabilities:

(i) P (A fails | B has failed)

(ii) P (A fails alone)

Solution:

(i) Let us assume the event which is failed by A is denoted by E

And the event which is failed by B is denoted by E

It is given in the question that,

The event failed by A, P (E_A) = 0.2

Event failed by both , P (E_A∩ E_B) = 0.15

And , event failed by alone = P ( E_B) – P (E_A∩E_B)

0.15 = P (EB) – 0.15

= P (E_B) = 0.30

Hence, P (E_A/E_B) = P (E_A∩E_B) /P (E_B)

= 0.15/0.3

= 0.5

(ii) We have , probability where A fails alone = P (E_A) – P (E_A∩E_B)

= 0.2 – 0.15

= 0.05

(ii) We have , the probability , where A fails alone

= P (E_A) – P (E_A∩E_B)

= 0.2 – 0.15

= 0.05

Question 10. Bag 1 contains 3 red and 4 black balls, and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II, and then a ball is drawn from Bag II. The ball so drawn is found to be red in color. Find the probability that the transferred ball is black.

Solution :

Let us first assume A1 denote the events that a red ball is transferred from bags I to II.

And A2 denote the event that a black ball is transferred from bags I to II.

∴ P (A₁) = 3/7

And, P (A₂) = 4/7

Let X be the event that the drawn ball is red

∴ when the red ball is transferred from bags I to II

P(X|A₁) = 5/10

= 1/2

And , when black ball is transferred from bag I to II ,

P(X|A₂) = 4/10

= 2/5

Hence , P (A₂|X) = P(A₂) P (X|A₂)\P(A₁)P(X|A₁) + P(A₂)P(X|A₂)

= 4/7 X 2/5/3/7 X 1/2 + 4/7 X 2/5

=16/31

Choose the correct answer in each of the following :

Question 11. A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then

A. A ⊂ B

B. B ⊂ A

C. B = φ

D. A = φ

Solution:

Option (A) is correct A ⊂ B

Question 12. If P (A|B) > P (A), then which of the following is correct.

A. P (B|A) < P (B)

B. P (A ∩ B) < P (A) . P (B)

C. P (B|A) > P (B)

D. P (B|A) = P (B)

Solution:

Option (C) is correct P (B|A) > P (B)

Question 13. If A and B are any two events such that P(A)+P(B) -P(A and B) =P(A) , then

A. P (B|A) =1

B. P(A|B) = 1

C. P(B|A) = 0

D. P(A|B) =0

Solution:

Option (B) is correct P(A|B) = 1