Exercise 11.5 - Chapter 11 Probability Distributions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 11.5$
Question 1.
Compute $\mathrm{P}(\mathrm{X}=\mathrm{k})$ for the binomial distribution, $\mathrm{B}(\mathrm{n}, \mathrm{p})$ where
(i) $n=6, p=\frac{1}{3}, k=3$
(ii) $n=10, p=\frac{1}{5}, k=4$
(iii) $n=9, p=\frac{1}{2}, k=7$
Solution:
(i) Given $n=6, p=\frac{1}{3}, k=3$
$
\begin{aligned}
\therefore q & =1-p=1-\frac{1}{3}=\frac{2}{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}=k) & =\mathrm{P}(\mathrm{X}=3) \\
& =6 \mathrm{C}_3\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^{6-3}=20\left(\frac{1}{27}\right)\left(\frac{8}{27}\right)=\frac{160}{729}
\end{aligned}
$
(ii) $n=10, p=\frac{1}{5}, k=4$
$
\begin{aligned}
\therefore q & =1-p=1-\frac{1}{5}=\frac{4}{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}=k) & =\mathrm{P}(\mathrm{X}=4) \\
& =10 \mathrm{C}_4\left(\frac{1}{5}\right)^4\left(\frac{4}{5}\right)^{10-4}=210\left(\frac{1}{5^4}\right)\left(\frac{4^6}{5^6}\right)=210\left(\frac{1}{5}\right)^4\left(\frac{4}{5}\right)^6
\end{aligned}
$
(iii) $n=9, p=\frac{1}{2}, k=7 ; q=1-p=1-\frac{1}{2}=\frac{1}{2}$
' $\mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots \ldots . n$
$\mathrm{P}(\mathrm{X}=k)=\mathrm{P}(\mathrm{X}=7)$
$
=9 C_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^{9-7}=36\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^2=36\left(\frac{1}{2}\right)^9
$

Question 2.
The probability that Mr. Q hits a target at any trial is $\frac{1}{4}$. Suppose he tries at the target 10 times. Find the probability that he hits the target
(i) exactly 4 times
(ii) at least one time.
Solution:
Let ' $p$ ' be the probability of hitting the trial
i.e., $p=\frac{1}{4}, \therefore q=1-p=1-\frac{1}{4}=\frac{3}{4}$
and number of trials ' $n$ ' $=10$
Probability of ' $x$ ' success in ' $n$ ' trials is
$\mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots \ldots n$
(i) Probability that Mr.Q hits the target exactly 4 times is
$
P(X=4)=10 C_4\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right)^{10-4}=10 C_4\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right)^6
$
(ii) Probability that Mr.Q hits the target atleast one time is
$
\mathrm{P}(\mathrm{X} \geq 1)=1-\mathrm{P}(\mathrm{X}<1)
$
$
=1-P(X=0)
$
$
\begin{aligned}
P(X \geq 1) & =1-P(X<1) \\
& =1-P(X=0) \\
& =1-10 C_0\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{10-0} \\
& =1-\left(\frac{3}{4}\right)^{10}
\end{aligned}
$

Question 3.
Using binomial distribution find the mean and variance of $\mathrm{X}$ for the following experiments
(i) A fair coin is tossed 100 times, and $\mathrm{X}$ denote the number of heads.
(ii) A fair die is tossed 240 times, and $\mathrm{X}$ denote the number of times that four appeared.
Solution:
(i) $\mathrm{n}=100$, ' $\mathrm{X}$ ' denotes the number of heads.
$\therefore$ Probability of getting a head $p=\frac{1}{2}$ and $q=1-p=\frac{1}{2}$
Mean $=n p=100 \times \frac{1}{2}=50$ Variance $=n p q=100 \times \frac{1}{2} \times \frac{1}{2}=25$
(ii) $\mathrm{n}=240$, ' $\mathrm{X}$ ' denotes the number of times four appeared.
$\therefore$ Probability of getting four $p=\frac{1}{6}$ and $q=1-p=\frac{5}{6}$
Mean $=n p=240 \times \frac{1}{6}=40 ;$ Variance $=n p q=240 \times \frac{1}{6} \times \frac{5}{6}=\frac{100}{3}$

Question 4.
The probability that a certain kind of component will survive a electrical test is $\frac{3}{4}$. Find the probability that exactly 3 of the 5 components tested survive.
Solution:
Given $\mathrm{n}=5$
Probability that a component survive in a test $=p=\frac{3}{4}$
$
\therefore q=1-p=1-\frac{3}{4}=\frac{1}{4}
$
Let ' $\mathrm{X}$ ' be the random variable denotes the number of components survived in a test.
Probability of ' $x$ ' successes in ' $n$ ' trials is.
$
\mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots \ldots n
$
Probability that exactly 3 components survive
$
\begin{aligned}
P(X=3) & =5 C_3\left(\frac{3}{4}\right)^3\left(\frac{1}{4}\right)^{5-3} \\
& =10\left(\frac{27}{64}\right)\left(\frac{1}{16}\right)=\frac{270}{1024}
\end{aligned}
$

Question 5.
A retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is $5 \%$. The inspector of the retailer randomly picks 10 items from a shipment. What is the probability that there will be
(i) at least one defective item
(ii) exactly two defective items.
Solution:
Given $\mathrm{n}=10$
Probability of a defective item $=p=5 \%=\frac{5}{100}$
$
\therefore q=1-p=1-\frac{5}{100}=\frac{95}{100}
$
Let ' $\mathrm{X}$ ' be the random variable denotes the number of defective items.

$\therefore$ Probability of ' $\mathrm{x}$ ' successes in ' $\mathrm{n}$ ' trials is
$
\mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots \ldots n
$
(i) Probability that atleast one defective item will be there
$
\begin{aligned}
\mathrm{P}(\mathrm{X} \geq 1) & =1-\mathrm{P}(\mathrm{X}<1) \\
& =1-[\mathrm{P}(\mathrm{X}=0)] \\
& =1-\left[10 \mathrm{C}_0\left(\frac{5}{100}\right)^0\left(\frac{95}{100}\right)^{10-0}\right] \\
& =1-\left(\frac{95}{100}\right)^{10}=1-(0.95)^{10}
\end{aligned}
$
(ii) Probability that exactly two defective item will be there
$
\begin{aligned}
P(X=2) & =10 C_2\left(\frac{5}{100}\right)^2\left(\frac{95}{100}\right)^8 \\
& =10 C_2(0.05)^2(0.95)^8
\end{aligned}
$

Question 6.
If the probability that a fluorescent light has a useful life of at least 600 hours is $0.9$, find the probabilities that among 12 such lights.
(i) exactly 10 will have a useful life of at least 600 hours;
(ii) at least 11 will have a useful life of at least 600 hours;
(iii) at least 2 will not have a useful life of at least 600 hours.
Solution:
Given $\mathrm{n}=12$
Probability that a fluorescent light has a life of atleast of 600 hours is $p=0.9$
$
\therefore \mathrm{q}=1-\mathrm{p}=1-0.9=0.1
$
Let ' $\mathrm{X}$ ' be the number of bulbs.
$\therefore$ Probability of ' $\mathrm{x}$ ' successes in ' $n$ ' trials is
$
\mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots \ldots . n
$
(i) Probability that exactly 10 bulbs will have a useful life of atleast 600 hours
$
\mathrm{P}(\mathrm{X}=10)=12 \mathrm{C}_{10}(0.9)^{10}(0.1)^2
$
(ii) Probability that atleast 11 will have a useful life of atleast 600 hours is
$
\begin{aligned}
\mathrm{P}(\mathrm{X} \geq 11) & =\mathrm{P}(\mathrm{X}=11)+\mathrm{P}(\mathrm{X}=12) \\
& =12 \mathrm{C}_{11}(0.9)^{11}(0.1)^1+12 \mathrm{C}_{12}(0.9)^{12}(0.1)^0 \\
& =12(0.9)^{11}(0.1)+(0.9)^{12}(1) \\
& =(0.9)^{11}(1.2+0.9)=(2.1)(0.9)^{11}
\end{aligned}
$

(iii) Probability that atleast 2 will not have a useful life of 600 hours is
$
\begin{aligned}
& =1-P(X \geq 2) \\
& =P(X<2) \\
& =P(X=0)+P(X=1) \\
& =12 \mathrm{C}_0(0.9)^0(0.1)^{12}+12 \mathrm{C}_1(0.9)^1(0.1)^{11} \\
& =(0.1)^{12}+12(0.9)(0.1)^{11} \\
& =(0.1)^{11}(0.1+10.8) \\
& =(10.9)(0.1)^{11}
\end{aligned}
$

Question 7.
The mean and standard deviation of a binomial variate $\mathrm{X}$ are respectively 6 and 2 . Find
(i) the probability mass function
(ii) $\mathrm{P}(\mathrm{X}=3)$
(iii) $\mathrm{P}(\mathrm{X} \geq 2)$.
Solution:
$\begin{aligned} \text { Given, Mean }= & n p=6 \\ \text { Standard deviation }= & \sqrt{n p q}=2 \\ & n p q=4\end{aligned}$
Substituting (1) in (2) $\Rightarrow \quad 6 q=4 \Rightarrow q=\frac{4}{6}=\frac{2}{3}$ $\therefore p=1-q=1-\frac{2}{3}=\frac{1}{3}$ $n p=6 \Rightarrow n\left(\frac{1}{3}\right)=6 \Rightarrow n=18$
Probability of ' $x$ ' success in ' $n$ ' trials is
$
\mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots \ldots n
$
(i) Probability mass function
$
\mathrm{P}(\mathrm{X}=x)=18 \mathrm{C}_x\left(\frac{1}{3}\right)^x\left(\frac{2}{3}\right)^{18-x}
$
(ii) $\mathrm{P}(\mathrm{X}=3)=18 \mathrm{C}_3\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^{15}$

(iii)
$
\begin{aligned}
P(X \geq 2) & =1-P(X<2) \\
& =1-[P(X=0)+P(X=1)] \\
& =1-\left[18 C_0\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^{18}+18 C_1\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^{17}\right] \\
& =1-\left[\left(\frac{2}{3}\right)^{18}+18\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{17}\right] \\
& =1-\left(\frac{2}{3}\right)^{17}\left[\frac{2}{3}+6\right]=1-\frac{20}{3}\left(\frac{2}{3}\right)^{17}
\end{aligned}
$

Question 8.
If $\mathrm{X} \sim \mathrm{B}(\mathrm{n}, \mathrm{p})$ such that $4 \mathrm{P}(\mathrm{X}=4)=\mathrm{P}(\mathrm{x}=2)$ and $\mathrm{n}=6$. Find the distribution, mean and standard deviation.
Solution:
$\mathrm{X} \sim \mathrm{B}(n, p)$ i.e., ' $\mathrm{X}$ ' is the binomial variate given $n=6$.
$
\begin{aligned}
\mathrm{P}(\mathrm{X}=x) & =n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots \ldots n \\
4 \mathrm{P}(\mathrm{X}=4) & =\mathrm{P}(\mathrm{X}=2) \\
4\left(6 \mathrm{C}_4 p^4 q^2\right) & =6 \mathrm{C}_2 p^4 q^4 \\
4 p^2 & =q^2 \\
4 p^2 & =(1-p)^2 \\
4 p^2-p^2+2 p-1 & =0 \\
3 p^2+2 p-1 & =0 \\
(3 p-1)(p+1) & =0 \\
\therefore p & =\frac{1}{3} ; p=-1 \text { is not possible. } \\
\text { If } p & =\frac{1}{3} \text { then } q=1-\frac{1}{3}=\frac{2}{3}
\end{aligned}
$
Binomial Distribution is $\mathrm{B}\left(6, \frac{1}{3}\right)$
$
\begin{aligned}
\text { Mean } n p & =6 \times \frac{1}{3}=2 \\
\text { Standard deviation } & =\sqrt{n p q}=\sqrt{6 \times \frac{1}{3} \times \frac{2}{3}} \\
& =\frac{2}{\sqrt{3}}=\frac{2 \sqrt{3}}{3}
\end{aligned}
$

Question 9.
In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are $0.4096$ and $0.2048$ respectively. Find the mean and variance of the distribution.
solution:
Number of trials $\mathrm{n}=5$
Probability of ' $\mathrm{x}$ ' successes in ' $n$ ' trials is
$
\mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots \ldots . n
$
Given $\mathrm{P}(\mathrm{X}=1)=0.4096$ and $\mathrm{P}(\mathrm{X}=2)=0.2048$
$
\begin{aligned}
& \mathrm{P}(\mathrm{X}=1)=0.4096 \Rightarrow 5 \mathrm{C}_1 p^1 q^4=5 p q^4=0.4096 \\
& \mathrm{P}(\mathrm{X}=2)=0.2048 \Rightarrow 5 \mathrm{C}_2 p^2 q^3=10 p^2 q^3=0.2048
\end{aligned}
$
Dividing Eq.(1) by Eq.(2)
$
\begin{array}{rlrl}
\frac{5 p q^4}{10 p^2 q^3} & =\frac{0.4096}{0.2048} \\
\frac{q}{2 p}=2 & \Rightarrow q=4 p & \\
& \Rightarrow 1-p=4 p & {[\because q=1-p]} \\
& \Rightarrow 5 p=1 &
\end{array}
$