Exercise 11.6 - Chapter 11 Probability Distributions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $11.6$
Choose the Correct or the most suitable answer from the given four alternatives:
Question 1.
Let $\mathrm{X}$ be random variable with probability density function
$
f(x)= \begin{cases}2 / x^3 & x \geq 1 \\ 0 & x<1\end{cases}
$
Which of the following statement is correct
(a) both mean and variance exist
(b) mean exists but variance does not exist
(c) both mean and variance do not exist
(d) variance exists but Mean does not exist
Solution:
(b) mean exists but variance does not exist
Hint:
Given probability density function $f(x)= \begin{cases}2 / x^3 & x \geq 1 \\ 0 & x<1\end{cases}$
$
\begin{aligned}
\text { Mean }=\mathrm{E}(\mathrm{X}) & =\int_{-\infty}^{\infty} x f(x) d x \\
& =\int_1^{\infty} x \cdot 2 / x^3 d x=\int_1^{\infty} 2 / x^2 d x \\
& =-(2 / x)_1^{\infty}=-(0-2)=2 \\
\mathrm{E}\left(\mathrm{X}^2\right) & =\int_1^{\infty} x^2 f(x) d x=\text { Undefined. As } \log \infty=\infty
\end{aligned}
$
Question 2 .
A rod of length 21 is broken into two pieces at random. The probability density function of the shorter of  the two pieces is $
f(x)= \begin{cases}\frac{1}{l}, & 0 $ The mean and variance of the shorter of the two pieces are respectively

(a) $\frac{l}{2}, \frac{l^2}{3}$
(b) $\frac{l}{2}, \frac{l^2}{6}$
(c) $l, \frac{l^2}{12}$
(d) $\frac{l}{2}, \frac{l^2}{12}$
Solution:
(d) $\frac{l}{2}, \frac{l^2}{12}$
Hint:
$
\begin{aligned}
f(x) & =\left\{\begin{array}{l}
\frac{1}{l}, \quad 0 0, \quad l \leq x<2 l
\end{array}\right. \\
\text { Mean }=\mathrm{E}(\mathrm{X}) & =\frac{1}{l} \int_0^l x d x=\frac{1}{l}\left[\frac{x^2}{2}\right]_0^l=\frac{1}{l}\left[\frac{l^2}{2}\right]=\frac{l}{2} \\
\mathrm{E}\left(\mathrm{X}^2\right) & =\frac{1}{l} \int_0^l x^2 d x=\frac{1}{l}\left[\frac{x^3}{3}\right]_0^l=\frac{1}{l}\left[\frac{l^3}{3}\right]=\frac{l^2}{3} \\
\operatorname{Variance}=\operatorname{Var}(\mathrm{X}) & =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& =\frac{l^2}{3}-\left(\frac{l}{2}\right)^2=\frac{l^2}{3}-\frac{l^2}{4}=\frac{l^2}{12}
\end{aligned}
$

Question 3.
Consider a game where the player tosses a six-sided fair die. If the face that comes up is 6 , the player wins $₹ 36$, otherwise he loses $₹ \mathrm{k}^2$, where $\mathrm{k}$ is the face that comes up $\mathrm{k}=\{1,2,3,4,5\}$. The expected amount to win at this game in ₹ is
(a) $\frac{19}{6}$
(b) $-\frac{19}{6}$
(c) $\frac{3}{2}$
(d) $-\frac{3}{2}$
Solution:
(b) $-\frac{19}{6}$
Hint:
Expected amount to win $=1-\frac{25}{6}=-\frac{19}{6}$
Question 4.
A pair of dice numbered 1,2,3,4,5,6 of a six-sided die and 1, 2,3,4 of a four-sided die is rolled and the sum is determined. Let the random variable $\mathrm{X}$ denote this sum. Then the number of elements in the
inverse image of 7 is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(d) 4
Hint:
Sample points to get the sum ' 7 ' are $(3,4),(4,3),(5,2),(6,1)$
Number of elements in inverse image of $7=4$

Question 5.
A random variable $\mathrm{X}$ has binomial distribution with $\mathrm{n}=25$ and $\mathrm{p}=0.8$ then standard deviation of $\mathrm{X}$ is
(a) 6
(b) 4
(c) 3
(d) 2
Solution:
(d) 2
Hint:
$\mathrm{n}=25, \mathrm{p}=0.8, \therefore \mathrm{q}=1-0.8=0.2$
Standard deviation $=\sqrt{n p q}=\sqrt{25 \times 0.8 \times 0.2}=\sqrt{4}=2$
Question 6.
Let $\mathrm{X}$ represent the difference between the number of heads and the number of tails obtained when a coin is tossed $\mathrm{n}$ times. Then the possible values of $\mathrm{X}$ are
(a) $i+2 n, i=0,1,2 \ldots n$
(b) $2 \mathrm{i}-\mathrm{n}, \mathrm{i}=0,1,2 \ldots \mathrm{n}$
(c) $\mathrm{n}-\mathrm{i}, \mathrm{i}=0,1,2 \ldots \mathrm{n}$
(d) $2 \mathrm{i}+2 \mathrm{n}, \mathrm{i}=0,1,2 \ldots \mathrm{n}$
Solution:
(b) $2 \mathrm{i}-\mathrm{n}, \mathrm{i}=0,1,2 \ldots \mathrm{n}$
Question 7.
If the function $f(x)=\frac{1}{12}$ for $a (a) 0 and 12
(b) 5 and 17
(c) 7 and 19
(d) 16 and 24
Solution:
(d) 16 and 24

Hint:
Since $f(x)$ is a probability density function, $\int_{-\infty}^{\infty} f(x) d x=1$
But when $a=16$ and $b=24, \int_{16}^{24} f(x) d x \neq 1$
Question 8.
Four buses carrying 160 students from the same school arrive at a football stadium. The buses carry, respectively, $42,36,34$, and 48 students. One of the students is randomly selected. Let $\mathrm{X}$ denote the number of students that were on the bus carrying the randomly selected student. One of the 4 bus drivers is also randomly selected. Let $\mathrm{Y}$ denote the number of students on that bus. Then $\mathrm{E}[\mathrm{X}]$ and $\mathrm{E}[\mathrm{Y}]$ respectively are
(a) 50,40
(b) 40,50
(c) $40.75,40$
(d) 41,41
Solution:
(c) $40.75,40$
Hint:
$\mathrm{E}[\mathrm{X}]>\mathrm{E}[\mathrm{Y}]$, Since the students selected is more likely to come from the bus carrying maximum number of students whereas the drivers are equally likely to be selected.
$
\begin{aligned}
& E[X]=\left(42 \times \frac{42}{160}\right)+\left(36 \times \frac{36}{160}\right)+\left(34 \times \frac{34}{160}\right)+\left(48 \times \frac{48}{160}\right)=40.75 \\
& E[Y]=\left(\frac{1}{4} \times 42\right)+\left(\frac{1}{4} \times 36\right)+\left(\frac{1}{4} \times 34\right)+\left(\frac{1}{4} \times 48\right)=40
\end{aligned}
$

Question 9.
Two coins are to be flipped. The first coin will land on heads with probability $0.6$, the second with Probability 0.5. Assume that the results of the flips are independent, and let $X$ equal the total number of heads that result. The value of $\mathrm{E}[\mathrm{X}]$ is $\ldots .$.
(a) $0.11$
(b) $1.1$
(c) 11
(d) 1
Solution:
(b) $1.1$
Hint:
$
\begin{aligned}
& \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{TT}) \\
& =\mathrm{P}(\mathrm{T}) \mathrm{P}(\mathrm{T})=(0.4)(0.5)=0.20 \\
& \mathrm{P}(\mathrm{X}=1)=\mathrm{P}(\mathrm{HT})+\mathrm{P}(\mathrm{TH})=(0.6)(0.5)+(0.4)(0.5)=0.5 \\
& \mathrm{P}(\mathrm{X}=2)=\mathrm{P}(\mathrm{HH}) \\
& =\mathrm{P}(\mathrm{H}) . \mathrm{P}(\mathrm{H})=(0.6)(0.5)=0.30 \\
& \mathrm{E}[\mathrm{X}]=\sum x \mathrm{P}(x) \\
& =(0) \cdot(0.20)+(1)(0.5)+(2)(0.30) \\
& =0.5+0.6=1.1
\end{aligned}
$

Question 10.
On a multiple-choice exam with 3 possible destructives for each of the 5 questions, the probability that a student will get 4 or more correct answers just by guessing is .......
(a) $\frac{11}{243}$
(b) $\frac{3}{8}$
(c) $\frac{1}{243}$
(d) $\frac{5}{243}$
Solution:
(a) $\frac{11}{243}$
Hint:
$
\begin{aligned}
p=\frac{1}{3}, q=\frac{2}{3} & \text { and } n=5 \\
\mathrm{P}(\mathrm{X}=x) & =n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots \ldots n \\
\mathrm{P}(\mathrm{X} \geq 4) & =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\
& =5 \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^1+5 \mathrm{C}_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^0 \\
& =5\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)+\left(\frac{1}{3}\right)^5=\left(\frac{1}{3}\right)^4\left[\frac{11}{3}\right]=\frac{11}{3^5}=\frac{11}{243}
\end{aligned}
$

Question 11.
If $\mathrm{P}\{\mathrm{X}=0\}=1-\mathrm{P}\{\mathrm{X}=1\}$. If $\mathrm{E}[\mathrm{X}]=3 \operatorname{Var}(\mathrm{X})$, then $\mathrm{P}\{\mathrm{X}=0\}$.
(a) $\frac{2}{3}$
(b) $\frac{2}{5}$
(c) $\frac{1}{5}$
(d) $\frac{1}{3}$
Solution:
(d) $\frac{1}{3}$
Hint:
Given $\mathrm{P}(\mathrm{X}=0)=1-\mathrm{P}(\mathrm{X}=1) \Rightarrow \mathrm{P}(\mathrm{X}=1)=1-\mathrm{P}(\mathrm{X}=0)$
Let $\mathrm{n}=\mathrm{P}(\mathrm{X}=0), \therefore \mathrm{P}(\mathrm{X}=1)=1-\mathrm{n}$
$\mathrm{E}(\mathrm{X})=(0)(\mathrm{n})+(1)(1-\mathrm{n})=1-\mathrm{n}$
$\mathrm{E}\left(\mathrm{X}^2\right)=(0)^2(\mathrm{n})+(1)^2(1-\mathrm{n})=1-\mathrm{n}$
Now substituting $\mathrm{E}(\mathrm{X})=3 \operatorname{Var}(\mathrm{X})$, we get $n=\frac{1}{3}$ (or) 1
$
\therefore \mathrm{P}(\mathrm{X}=0)=\frac{1}{3}
$
Question 12.
If $X$ is a binomial random variable with expected value 6 and variance $2.4$, Then $P\{X=5\}$ is
(a) $\left(\begin{array}{c}10 \\ 5\end{array}\right)\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4$
(b) $\left(\begin{array}{c}10 \\ 5\end{array}\right)\left(\frac{3}{5}\right)^5$
(c) $\left(\begin{array}{c}10 \\ 5\end{array}\right)\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^6$
(d) $\left(\begin{array}{c}10 \\ 5\end{array}\right)\left(\frac{3}{5}\right)^5\left(\frac{2}{5}\right)^5$
Solution:
(d) $\left(\begin{array}{c}10 \\ 5\end{array}\right)\left(\frac{3}{5}\right)^5\left(\frac{2}{5}\right)^5$
Hint:
Given, $n p=6$ and $n p q=2.4$
$
\begin{aligned}
q=\frac{n p q}{n p} & =\frac{2.4}{6}=0.4 \quad \therefore p=1-q=1-0.4=0.6 \\
n p & =6 \\
n(0.6) & =6 \Rightarrow n=\frac{6}{0.6}=10 \\
\mathrm{P}(\mathrm{X}=5) & =10 \mathrm{C}_5\left(\frac{6}{10}\right)^5\left(\frac{4}{10}\right)^5=10 \mathrm{C}_5\left(\frac{3}{5}\right)^5\left(\frac{2}{5}\right)^5
\end{aligned}
$

Question 13.
The random variable $\mathrm{X}$ has the probability density function $f(x)=\left\{\begin{array}{cc}a x+b, & 0 (a) 1 and $\frac{1}{2}$
(b) $\frac{1}{2}$ and 1
(c) 2 and 1
(d) 1 and 2
Solution:
(a) 1 and $\frac{1}{2}$

Hint:
Given probability density function $f(x)=\left\{\begin{array}{cc}a x+b, & 0 Since, $f(x)$ is a probability density function, $\int_0^1(a x+b) d x=1$
$
\begin{aligned}
{\left[\frac{a x^2}{2}+b x\right]_0^1 } & =1 \\
\frac{a}{2}+b & =1 \Rightarrow a+2 b=2 \\
\mathrm{E}(\mathrm{X})=\frac{7}{12} & \Rightarrow \int_0^1\left(a x^2+b x\right) d x=\frac{7}{12} \\
{\left[\frac{a x^3}{3}+\frac{b x^2}{2}\right]_0^1 } & =\frac{7}{12} \\
\frac{a}{3}+\frac{b}{2} & =\frac{7}{12} \Rightarrow 2 a+3 b=\frac{7}{2}
\end{aligned}
$
Solving these equations, we get $a=1$ and $b=\frac{1}{2}$
Question 14.
Suppose that $\mathrm{X}$ takes on one of the values 0,1 , and 2 . If for some constant $\mathrm{k}$, $P(X=i)=k P(X=i-1)$ for $i=1,2 P(X=0)=\frac{1}{7}$. Then the value of $k$ is .....
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2
Hint:
$\mathrm{P}(\mathrm{X}=\mathrm{i})=\mathrm{k} P(\mathrm{X}=\mathrm{i}-1)$ for $\mathrm{i}=1,2$ and $\mathrm{P}(\mathrm{X}=0)=\frac{1}{7}$
Solving the equation, we get $\mathrm{k}=2$
Question 15.
Which of the following is a discrete random variable?
I. The number of cars crossing a particular signal in a day.
II. The number of customers in a queue to buy train tickets at a moment.
III. The time taken to complete a telephone call.
(a) I and II
(b) II only

(c) III only
(d) II and III
Solution:
(a) I and II

Question 16.
$f(x)= \begin{cases}2 x, & 0 \leq x \leq a \\ 0, & \text { otherwise }\end{cases}$ is a probability density function of a random variable, then the value of a is .......
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(a) 1
Hint:
Since, $f(x)$ is a probability density function, $\int_0^a f(x) d x=1$
$
\begin{aligned}
& \Rightarrow 2 \int_0^a x d x=1 \Rightarrow\left[x^2\right]_0^a=1 \\
& a^2=1 \Rightarrow a=1
\end{aligned}
$

Question 17.
The probability function of a random variable is defined as ........

Then $\mathrm{E}(\mathrm{X})$ is equal to ......
(a) $\frac{1}{15}$
(b) $\frac{1}{10}$
(c) $\frac{1}{3}$
(d) $\frac{2}{3}$
Solution:
(d) $\frac{2}{3}$
Hint:
$
\begin{array}{rl|rl}
\mathrm{E}(\mathrm{X}) & =\Sigma x f(x) & \Sigma \mathrm{P}_i & =1 \\
\mathrm{E}(\mathrm{X}) & =-2 k-2 k+0+4 k+10 k & \therefore 15 k & =1 \\
& =10 k=\frac{10}{15}=\frac{2}{3} & k & =\frac{1}{15}
\end{array}
$

Question 18.
Let $\mathrm{X}$ have a Bernoulli distribution with mean $0.4$, then the variance of $(2 \mathrm{X}-3)$ is $\ldots .$.
(a) $0.24$
(b) $0.48$
(c) $0.6$
(d) $0.96$
Solution:
(d) $0.96$
Hint:
$
\begin{aligned}
& \text { Mean }=0.4 \Rightarrow \mu=p=0.4 \\
& \operatorname{Var}(2 \mathrm{X}-3)=2^2 \operatorname{Var} \mathrm{X} \\
& =4(\mathrm{pq})=4(0.4)(0.6)=0.96
\end{aligned}
$
Question 19.
If in 6 trials, $X$ is a binomial variate which follows the relation $9 P(X=4)=P(X=2)$, then the probability of success is ....
(a) $0.125$
(b) $0.25$
(c) $0.375$
(d) $0.75$
Solution:
(b) $0.25$
Hint:
$
n=6, \quad \mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots \ldots . n
$
Given, 9P $(X=4)=P(X=2)$
$
\begin{aligned}
& 96 \mathrm{C}_4 p^4 q^2=6 \mathrm{C}_2 p^2 q^4 \\
& 9 \mathrm{p}^2=\mathrm{q}^2\left[\because 6 \mathrm{C}_4=6 \mathrm{C}_2\right] \\
& 9 \mathrm{p}^2=(1-\mathrm{p})^2 \\
& 9 \mathrm{p}^2=1-2 \mathrm{p}+\mathrm{p}^2 \\
& 8 \mathrm{p}^2+2 \mathrm{p}-1=0 \\
&(4 \mathrm{p}-1)(2 \mathrm{p}+1)=0 \\
& p=\frac{1}{4},-\frac{1}{2} \quad \quad\left[p=-\frac{1}{2} \text { is not possible }\right] \\
& p=\frac{1}{4}=0.25 \quad
\end{aligned}
$

Question 20.
A computer salesperson knows from his past experience that he sells computers to one in every twenty customers who enter the showroom. What is the probability that he will sell a computer to exactly two of the next three customers?
(a) $\frac{57}{20^3}$
(b) $\frac{57}{20^2}$
(c) $\frac{19^3}{20^3}$.
(d) $\frac{57}{20}$
Solution:
(a) $[1$ taex $] \backslash \operatorname{frac}\{57\}\left\{20^{\wedge}\{3\}\right\}[/$ latex $]$
Hint:
$
\begin{aligned}
& p=\frac{1}{20}, q=\frac{19}{20} \text { and } n=3 \\
& \mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots \ldots n \\
& \mathrm{P}(\mathrm{X}=2)=3 \mathrm{C}_2\left(\frac{1}{20}\right)^2\left(\frac{19}{20}\right)^1=3\left(\frac{1}{20^2}\right)\left(\frac{19}{20}\right)=\frac{57}{(20)^3}
\end{aligned}
$