Miscellaneous Exercise (Revised) - Chapter 13 - Probability - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions Chapter 13: Probability

Miscellaneous Exercise Question 1.

A and $B$ are two events such that $P(A) \neq 0$. Find $P(B / A)$ if:
(i)A is a subset of $B$
(ii) $\mathbf{A} \cap \mathbf{B}=\phi$

Answer.

A and B are two events such that $P(A) \neq 0$
To find: $P(B \mid A)$
(i) $\mathrm{A}$ is a subset of $\mathrm{B}$
$
\begin{aligned}
& \Rightarrow A \subset B \\
& P(A \cap B)=P(A) \\
& \therefore P(B \mid A)=\frac{P(A \cap B)}{P(A)}=\frac{P(A)}{P(A)}=1
\end{aligned}
$
(ii) $\mathrm{A} \cap \mathrm{B}=\phi$
$
\therefore P(B \mid A)=\frac{P(A \cap B)}{P(A)}=\frac{0}{P(A)}=0
$
Miscellaneous Exercise Question2.

A couple has two children.
(i)Find the probability that both children are males, 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.

Answer.

(i) Let $B_1$ and $G_1$ stand for male and female respectively.

Now the sample space is $(S)=\left\{B_1 B_2, B_1 G_2, B_2 G_1, G_1 G_2\right\}$

Let us consider the following events,
$
\begin{aligned}
& A=\text { both are males } \\
& B=\text { at least one is a male } \\
& \therefore A=\left\{B_1 B_2\right\} \text { and } B=\left\{B_1 B_2, B_1 G_2, B_2 G_1\right\} \\
& P(B)=\frac{3}{4}, A \cap B=\left\{B_1 B_2\right\} \text { and } P(A \cap B)=\frac{1}{4} \\
& \therefore \text { Required probability }=P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 4}{3 / 4}=\frac{1}{3}
\end{aligned}
$
(ii) Let $A=$ both are females, $A=\left\{G_1 G_2\right\}$ and $C=$ the older is a girl
$
\begin{aligned}
& \mathrm{C}=\left\{\mathrm{B}_1 \mathrm{~B}_2, \mathrm{~B}_1 \mathrm{G}_2\right\} \Rightarrow \mathrm{P}(\mathrm{C})=\frac{2}{4} \\
& \therefore \text { Required probability }=\mathrm{P}(\mathrm{A} \mid \mathrm{C})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{C})}{\mathrm{P}(\mathrm{C})}=\frac{1 / 4}{2 / 4}=\frac{1}{2}
\end{aligned}
$

Miscellaneous Exercise 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 are equal number of males and females.

Answer.

Men are represented by $\mathrm{E}_1$ and women are represented by $\mathrm{E}_2$.
$
\therefore \mathrm{P}\left(\mathrm{E}_1\right)=\frac{1}{2} \text { and } \mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{2}
$

A represents grey hair persons.

$\begin{aligned}
& \therefore P\left(A \mid E_1\right)=\frac{5}{100} \text { and } P\left(A \mid E_2\right)=\frac{25}{10000} \\
& P\left(E_1 \mid A\right)=\frac{P\left(E_1\right) P\left(A \mid E_1\right)}{P\left(E_1\right) P\left(A \mid E_1\right)+P\left(E_2\right) P\left(A \mid E_2\right)} \\
& =\frac{\frac{1}{2} \times \frac{5}{100}}{\frac{1}{2} \times \frac{5}{100}+\frac{1}{2} \times \frac{25}{10000}}=\frac{\frac{5}{100}}{\frac{5}{100}+\frac{25}{10000}}=\frac{\frac{5}{100}}{\frac{525}{10000}}=\frac{5}{100} \times \frac{10000}{525}=\frac{20}{21}
\end{aligned}$

Miscellaneous Exercise Question 4.

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

Answer.

$p=\frac{90}{100}=\frac{9}{10}$ and $q=1-p=1-\frac{9}{10}=\frac{1}{10}$
$
\begin{aligned}
& n=10 \\
& \mathrm{P}(\text { at most } 6 \text { successes })=1-[\mathrm{P}(\mathrm{X}=7)+\mathrm{P}(\mathrm{X}=8)+\mathrm{P}(\mathrm{X}=9)+\mathrm{P}(\mathrm{X}=10)] \\
& =1-\left[{ }^{10} \mathrm{C}_7\left(\frac{9}{10}\right)^7\left(\frac{1}{10}\right)^3+{ }^{10} \mathrm{C}_8\left(\frac{9}{10}\right)^8\left(\frac{1}{10}\right)^2+{ }^{10} \mathrm{C}_9\left(\frac{9}{10}\right)^9\left(\frac{1}{10}\right)+{ }^{10} \mathrm{C}_{10}\left(\frac{9}{10}\right)^{10}\right] \\
& =1-\sum_{r=7}^{10}{ }^{10} \mathrm{C}_r(0.9)^\gamma(0.1)^{10-\gamma}
\end{aligned}
$
Miscellaneous Exercise Question 5.

If a leap year is selected at random, what is the change that it will contain 52 Tuesday?

Answer.

A leap year has 366 days which means 52 complete weeks and 2 days. If any one of these two days in a Tuesday then the year will have 53 Tuesdays.

Number of total days in a week $=7$
Number of favourable days $=2$

Therefore, P (the year will have 53 Tuesday) $=\frac{2}{7}$
Miscellaneous Exercise Question 6

Suppose we have four boxes $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ 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 $\mathrm{A}$ ?, box $\mathrm{B}$ ? box $\mathrm{C}$ ?

Answer.

Let $\mathrm{R}$ represents the drawing of red ball and the four boxes are represented by A, B, C and D.
$
\begin{aligned}
& \operatorname{So} P(R \mid A)=\frac{1}{10}, P(R \mid B)=\frac{6}{10} \\
& P(R \mid C)=\frac{8}{10}, P(R \mid D)=\frac{0}{10}=0
\end{aligned}
$

Since there are 4 bags.
Therefore, $\mathrm{P}(\mathrm{A})=\frac{1}{4}, \mathrm{P}(\mathrm{B})=\frac{1}{4}, \mathrm{P}(\mathrm{C})=\frac{1}{4}, \mathrm{P}(\mathrm{D})=\frac{1}{4}$
$
\begin{aligned}
& \mathrm{P}(\mathrm{A} \mid \mathrm{R})=\frac{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{R} \mid \mathrm{A})}{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{R} \mid \mathrm{A})+\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{R} \mid \mathrm{B})+\mathrm{P}(\mathrm{C}) \cdot \mathrm{P}(\mathrm{R} \mid \mathrm{C})+\mathrm{P}(\mathrm{D}) \cdot \mathrm{P}(\mathrm{R} \mid \mathrm{D})} \\
& =\frac{\frac{1}{4} \times \frac{1}{10}}{\frac{1}{4} \times \frac{1}{10}+\frac{1}{4} \times \frac{6}{10}+\frac{1}{4} \times \frac{8}{10}+\frac{1}{4} \times 0} \\
& =\frac{\frac{1}{10}}{\frac{1}{10}+\frac{6}{10}+\frac{8}{10}}=\frac{1}{15} \\
& P(B \mid R)=\frac{P(B) \cdot P(R \mid B)}{P(A) P(R \mid A)+P(B) P(R \mid B)+P(C) P(R \mid C)+P(D) P(R \mid D)} \\
& =\frac{\frac{1}{4} \times \frac{6}{10}}{\frac{1}{4} \times \frac{1}{10}+\frac{1}{4} \times \frac{6}{10}+\frac{1}{4} \times \frac{8}{10}+\frac{1}{4} \times 0}=\frac{6}{15}=\frac{2}{5} \\
&
\end{aligned}
$

$P(C \mid R)=\frac{P(C) P(R \mid C)}{P(A) P(R \mid A)+P(B) P(R \mid B)+P(C) P(R \mid C)+P(D) P(R \mid D)}$

$=\frac{\frac{1}{4} \times \frac{8}{10}}{\frac{1}{4} \times \frac{1}{10}+\frac{1}{4} \times \frac{6}{10}+\frac{1}{4} \times \frac{8}{10}+\frac{1}{4} \times 0}=\frac{8}{15}$

Miscellaneous Exercise Question 13.

Assume that the chances of a patient having a heart attack is $40 \%$. It is also assumed that a meditation and yoga course reduces the risk of heart attack by $30 \%$ and prescription of 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.

Answer.

A patient has options to have the treatment of yoga and meditation and that of prescription of drugs.

Let these events be denoted by $\mathrm{E}_1$ and $\mathrm{E}_2$ i.e.,
$\mathrm{E}_1=$ Treatment of yoga and meditation
$\mathrm{E}_2=$ Treatment of prescription of certain drugs
$\mathrm{P}\left(\mathrm{E}_1\right)=\frac{1}{2}$ and $\mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{2}$
Let $\mathrm{A}$ denotes that a person has heart attack, then $\mathrm{P}(\mathrm{A})=40 \%=0.40$
Yoga and meditation reduces heart attack by 30.
$\Rightarrow$ Inspite of getting yoga and meditation treatment heart risk is $70 \%$ of 0.40
$\Rightarrow P\left(A \mid E_1\right)=0.40 \times 0.70=0.28$
Also, Drug prescription reduces the heart attack rick by $25 \%$
Even after adopting the drug prescription hear rick is $75 \%$ of 0.40
$
\Rightarrow \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_2\right)=0.40 \times 0.75=0.30
$

$\begin{aligned}
& P\left(E_1 \mid A\right)=\frac{P\left(E_1\right) P\left(A \mid E_1\right)}{P\left(E_1\right) P\left(A \mid E_1\right)+P\left(E_2\right) P\left(A \mid E_2\right)} \\
& =\frac{\frac{1}{2} \times 0.28}{\frac{1}{2} \times 0.28+\frac{1}{2} \times 0.30} \\
& =\frac{0.28}{0.28+0.30}=\frac{0.28}{0.58}=\frac{28}{58}=\frac{14}{29}
\end{aligned}$

Miscellaneous Exercise Question 14.

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

Answer.

There are four entries in a determinant of $2 \times 2$ order. Each entry may be filled up in two ways with 0 or 1.
$\therefore$ Number of determinants that can be formed $=2^4=16$
The value of determinants is positive in the following cases:
$
\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|=\quad\left|\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right|=\quad\left|\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right|=3
$

Therefore, the probability that the determinant is positive $=\frac{3}{16}$
Miscellaneous Exercise Question 15.

An electronic assembly consists of two sub-systems say $A$ and $B$. From previous testing procedures, the following probabilities are assumed to be known:
$
\begin{aligned}
& P(A \text { fails })=0.2 \\
& P(B \text { fails alone })=0.15
\end{aligned}
$

$P(A$ and $B$ fail $)=0.15$
Evaluate the following probabilities.
Answer.

Event A fails and B fails denoted by $\overline{\mathrm{A}}$ and $\overline{\mathrm{B}}$ respectively.
$
\begin{aligned}
& \therefore P(\bar{A})=0.2 \text { and } P(A \text { and } B \text { fails })=0.15 \\
& \Rightarrow P(A \cap B)=0.15 \\
& \therefore P(\bar{B} \text { above })=P(\bar{B})-P(A \cap B) \\
& \Rightarrow 0.15=P(\bar{B})-0.15 \\
& \Rightarrow P(\bar{B})=0.30
\end{aligned}
$
(i) $\mathrm{P}(\overline{\mathrm{A}} \mid \overline{\mathrm{B}})=\frac{\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})}{\mathrm{P}(\overline{\mathrm{B}})}=\frac{0.15}{0.30}=\frac{1}{2}=0.5$
(ii) $\mathrm{P}(\mathrm{A}$ fails alone $)=\mathrm{P}(\overline{\mathrm{A}}$ alone $)=\mathrm{P}(\overline{\mathrm{A}})-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=0.20-0.15=0.05$

Miscellaneous Exercise Question 16.

Bag I contains 3 Red and 4 Black balls and B II contains 4 Red and \% 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 $\mathrm{E}_1=$ Ball transferred from Bag I to Bag II is red
$\mathrm{E}_2=$ Ball transferred from Bag I to Bag Ii is black
$A=$ Ball drawn from Bag II is red in colour
$
P\left(E_1\right)=\frac{3}{7} \text { and } P\left(E_2\right)=\frac{4}{7}
$

$
\begin{aligned}
& \Rightarrow P\left(A \mid E_1\right)=\frac{5}{10}=\frac{1}{2} \text { and } P\left(A \mid E_2\right)=\frac{4}{10}=\frac{2}{5} \\
& P\left(E_2 \mid A\right)=\frac{P\left(E_2\right) \cdot P\left(A \mid E_2\right)}{P\left(E_1\right) \cdot P\left(A \mid E_1\right)+P\left(E_2\right) \cdot P\left(A \mid E_2\right)} \\
& =\frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2}+\frac{4}{7} \times \frac{2}{5}}=\frac{16}{31}
\end{aligned}
$

Choose the correct answer in each of the following:
Miscellaneous Exercise Question 17.

If $A$ and $B$ are two events such that $P(A) \neq 0$ and $P(B \mid A)=1$, then:
(A) $\mathrm{A} \subset \mathrm{B}$
(B) $\mathbf{B} \subset \mathbf{A}$
(C) $\mathbf{B}=\phi$
(D) $\mathrm{A}=\phi$

Answer.

$A$ and $B$ are two events such that $P(A) \neq 0$ and $P(B \mid A)=1$
$
\begin{aligned}
& \Rightarrow \frac{P(A \cap B)}{P(A)}=1 \Rightarrow P(A \cap B)=P(A) \\
& \Rightarrow(A \cap B)=(A) \Rightarrow A \subset B
\end{aligned}
$

Therefore, option (A) is correct.
Miscellaneous Exercise Question 18.

If $\mathbf{P}(A \mid B)>P(A)$, then which of the following is correct:

(A) $\mathbf{P}($ B $\mid$ A $)<\mathbf{P}$ (B)
(B) $\mathbf{P}(\mathrm{A} \cap \mathrm{B})<\mathbf{P}$ (A) . P (B)

(C) $\mathbf{P}(B \mid A)>\mathbf{P}$ (B)
(D) $\mathbf{P}(B \mid A)=\mathbf{P}(\mathbf{B})$

Answer.

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

Therefore, option (C) is correct.
Miscellaneous Exercise Question 19.

If $A$ and $B$ are any two events such that $P(A)+P(B)-P(A$ and $B)=P(A)$, then
(A) $\mathbf{P}(B \mid A)=\mathbf{1}$
(B) $\mathbf{P}(A \mid B)=\mathbf{1}$
(C) $\mathbf{P}(B \mid A)=0$
(D) $\mathbf{P}(A \mid B)=\mathbf{0}$

Answer.

$P(A)+P(B)-P(A$ and $B)=P(A)$
$
\begin{aligned}
& \Rightarrow P(B)-P(A \text { and } B) \\
& \Rightarrow \frac{P(A \cap B)}{P(B)}=1 \\
& \Rightarrow P(A \mid B)=1
\end{aligned}
$

Therefore, option (B) is correct.