NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 12 - Probability

Question 1(i): A and B are two events such that$P(A)\ne 0.$ Find $P(B\mid A),$ if

$A$ is a subset of $B$

Answer:

A and B are two events such that$P(A)\ne 0.$

$A\subset B$

$\Rightarrow A\cap B=A$

$P(A\cap B)=P(B\cap A)=P(A)$

$P(B|A)=\frac{P(B\cap A)}{P(A)}$

$P(B|A)=\frac{P(A)}{P(A)}$

$P(B|A)=1$

Question 1(ii): $A$ and $B$ are two events such that $P(A)\ne 0.$ Find $P(B\mid A),$ if

$A\cap B=\varphi$

Answer:

A and B are two events such that$P(A)\ne 0.$

$P(A\cap B)=P(B\cap A)=0$

$P(B|A)=\frac{P(B\cap A)}{P(A)}$

$P(B|A)=\frac{0}{P(A)}$

$P(B|A)=0$

Question 2 (i): A couple has two children,

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

Answer:

A couple has two children,

sample space $=\{(b,b),(g,g),(b,g),(g,b)\}$

Let A be both children are males and B is at least one of the children is male.

$(A\cap B)=\{(b,b)\}$

$P(A\cap B)=\frac{1}{4}$

$P(A)=\frac{1}{4}$

$P(B)=\frac{3}{4}$

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

$P(A|B)=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

Question 2 (ii): A couple has two children,

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

Answer:

A couple has two children,

sample space $=\{(b,b),(g,g),(b,g),(g,b)\}$

Let A be both children are females and B be the elder child is a female.

$(A\cap B)=\{(g,g)\}$

$P(A\cap B)=\frac{1}{4}$

$P(A)=\frac{1}{4}$

$P(B)=\frac{2}{4}$

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

$P(A|B)=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2}$

Question 3: Suppose that $5^o{/}_o$ of men and ${0.25}^o{/}_o$ 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 are equal number of males and females.

Answer:

We have $5^o{/}_o$ of men and ${0.25}^o{/}_o$ of women have grey hair.

Percentage of people with grey hairs $=(5+0.25)\mathrm{\%}=5.25\mathrm{\%}$

The probability that the selected haired person is male :

$=\frac{5}{5.25}=\frac{20}{21}$

Question 4: Suppose that ${90}^o{/}_o$ of people are right-handed. What is the probability that at most $6$ of a random sample of $10$ people are right-handed?

Answer:

${90}^{\circ }\mathrm{\%}$ of people are right-handed.

$\begin{array}{rl}& P(\text{ right }-\text{ handed })=\frac{9}{10}\\ & P(\text{ left }-\text{ handed })=q=1-\frac{9}{10}=\frac{1}{10}\end{array}$

at most 6 of a random sample of 10 people are right-handed.
the probability that more than 6 of a random sample of 10 people are right-handed is given by,

$\begin{array}{rl}& {}^{10}{C}_r{P}^r{q}^{10-r}\\ & =\sum _T^{1010}{C}_r{\frac{9}{10}}^r\cdot {\left(\frac{1}{10}\right)}^{10-r}\end{array}$

the probability that at most 6 of a random sample of 10 people are right-handed is given by

$=1-\sum _T^{1010}{C}_r\cdot {\frac{9}{10}}^r\cdot {\left(\frac{1}{10}\right)}^{10-r}$

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

Answer:

In a leap year, there are 366 days.

In 52 weeks, there are 52 Tuesdays.

The probability that a leap year will have 53 Tuesday is equal to the probability that the remaining 2 days are Tuesday.

The remaining 2 days can be :

1. Monday and Tuesday

2. Tuesday and Wednesday

3. Wednesday and Thursday

4. Thursday and Friday

5.friday and Saturday

6.saturday and Sunday

7.sunday and Monday

Total cases = 7.

Favorable cases = 2

Probability of having 53 Tuesday in a leap year = P.

$P=\frac{2}{7}$

Question 6 (i): Suppose we have four boxes A,B,C and D containing coloured marbles as given below:

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 ?

Answer:

'

Let R be the event of drawing red marble.

Let $E_A,E_B,E_C$ respectively denote the event of selecting box A, B, C.

Total marbles = 40

Red marbles =15

$P(R)=\frac{15}{40}=\frac{3}{8}$

Probability of drawing red marble from box A is $P(E_A|R)$

$P(E_A|R)=\frac{P(E_A\cap R)}{P(R)}$

$=\frac{\frac{1}{40}}{\frac{3}{8}}$

$=\frac{1}{15}$

Question 6 (ii): Suppose we have four boxes A,B,C and D containing coloured marbles as given below:

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 B?

Answer:

Let R be event of drawing red marble.

Let $E_A,E_B,E_C$ respectivly denote event of selecting box A,B,C.

Total marbles = 40

Red marbles =15

$P(R)=\frac{15}{40}=\frac{3}{8}$

Probability of drawing red marble from box B is $P(E_B|R)$

$P(E_B|R)=\frac{P(E_B\cap R)}{P(R)}$

$=\frac{\frac{6}{40}}{\frac{3}{8}}$

$=\frac{2}{5}$

Question 6 (iii): Suppose we have four boxes A,B,C and D containing coloured marbles as given below:

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 C?

Answer:

Let R be event of drawing red marble.

Let $E_A,E_B,E_C$ respectivly denote event of selecting box A,B,C.

Total marbles = 40

Red marbles =15

$P(R)=\frac{15}{40}=\frac{3}{8}$

Probability of drawing red marble from box C is $P(E_C|R)$

$P(E_C|R)=\frac{P(E_C\cap R)}{P(R)}$

$=\frac{\frac{8}{40}}{\frac{3}{8}}$

$=\frac{8}{15}$

Question 7: Assume that the chances of a patient having a heart attack is ${40}^o{/}_o.$ It is also assumed that a meditation and yoga course reduce the risk of heart attack by ${30}^o{/}_o$ and prescription of certain drug reduces its chances by ${25}^o{/}_o.$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?

Answer:

Let A,E1, E2 respectively denote the event that a person has a heart break, selected person followed the course of yoga and meditation , and the person adopted

the drug prescription.

$\therefore P(A)=0.40$

$\therefore P(E1)=P(E2)=\frac{1}{2}$

$P(A|E1)=0.40\times 0.70=0.28$

$P(A|E2)=0.40\times 0.75=0.30$

the probability that the patient followed a course of meditation and yoga is $P(E1,A)$

$P(E1,A)=\frac{P(E1).P(E1|A)}{P(E1).P(E1|A)+P(E2).P(E2|A)}$

$P(E1,A)=\frac{0.5\times 0.28}{0.5\times 0.28+0.5\times 0.30}$

$=\frac{14}{29}$

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 $\frac{1}{2}$ ).

Answer:

Total number of determinant of second order with each element being 0 or 1 is $2^4=16$

The values of determinant is positive in the following cases $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right],\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$

Probability is

$=\frac{3}{16}$

Question 9 (i): 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

$P(Afails\mid Bhasfailed)$

Answer:

Let event in which A fails and B fails be $E_A,E_B$

$P(E_A)=0.2$

$P(E_{A}and{E}_B)=0.15$

$P(Bfailsalone)=P(E_B)-P(E_{A}and{E}_B)$

$\Rightarrow 0.15=P(E_B)-0.15$

$\Rightarrow P(E_B)=0.3$

$P(E_A|E_B)=\frac{P(E_A\cap E_B)}{P(E_B)}$

$=\frac{0.15}{0.3}=0.5$

Question 9 (ii): 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

$P(Afailsalone)$

Answer:

Let event in which A fails and B fails be $E_A,E_B$

$P(E_A)=0.2$

$P(E_{A}and{E}_B)=0.15$

$P(Bfailsalone)=P(E_B)-P(E_{A}and{E}_B)$

$\Rightarrow 0.15=P(E_B)-0.15$

$\Rightarrow P(E_B)=0.3$

$P(Afailsalone)=P(E_A)-P(E_{A}and{E}_B)$

$=0.2-0.15=0.05$

Question 10: Bag I 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 colour. Find the probability that the transferred ball is black.

Answer:

Let E1 and E2 respectively denote the event that red ball is transfered from bag 1 to bag 2 and a black ball is transfered from bag 1 to bag2.

$P(E1)=\frac{3}{7}$ and $P(E2)=\frac{4}{7}$

Let A be the event that ball drawn is red.

When a red ball is transfered from bag 1 to bag 2.

$P(A|E1)=\frac{5}{10}=\frac{1}{2}$

When a black ball is transfered from bag 1 to bag 2.

$P(A|E2)=\frac{4}{10}=\frac{2}{5}$

$P(E2|A)=\frac{P(E2).P(A|E2)}{P(E2).P(A|E2)+P(E1).P(A|E1)}$

$=\frac{\frac{4}{7}\times \frac{2}{5}}{\frac{4}{7}\times \frac{2}{5}+\frac{3}{7}\times \frac{1}{2}}$

$=\frac{16}{31}$

Question 11: If A and B are two events such that $P(A\ne 0)$ and$P(B\mid A)=1,$ then

Choose the correct answer of the following:

(A) $A\subset B$

(B) $B\subset A$

(C) $B=\varphi$

(D) $A=\varphi$

Answer:

A and B are two events such that $P(A\ne 0)$ and$P(B\mid A)=1,$

$P(B|A)=\frac{P(B\cap A)}{P(A)}$

$1=\frac{P(B\cap A)}{P(A)}$

$P(B\cap A)=P(A)$

$\Rightarrow A\subset B$

Option A is correct.

Question 12: If $P(A\mid B)>P(A)$, then which of the following is correct :

(A) $P(B\mid A)<P(B)$

(B) $P(A\cap B)<P(A).P(B)$

(C) $P(B\mid A)>P(B)$

(D)$P(B\mid A)=P(B)$

Answer:

$P(A\mid B)>P(A)$

$\Rightarrow \frac{P(A\cap B)}{P(B)}>P(A)$

$\Rightarrow P(A\cap B)>P(A).P(B)$

$\Rightarrow \frac{P(A\cap B)}{P(A)}>P(B)$

$\Rightarrow P(B|A)>P(B)$

Option C is correct.

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

(A) $P(B\mid A)=1$

(B) $P(A\mid B)=1$

(C) $P(B\mid A)=0$

(D) $P(A\mid B)=0$

Answer:

$P(A)+P(B)-P(AandB)=P(A),$

$P(A)+P(B)-P(A\cap B)=P(A)$

$\Rightarrow P(B)-P(A\cap B)=0$

$\Rightarrow P(B)=P(A\cap B)$

$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B)}{P(B)}=1$

Option B is correct.