• Let $F$ be $a$ set with $(2 m+1)$ elements. Let $\varepsilon$ be the class of all subsets of $F$ with odd number of elements. A set is randomly selected from $\varepsilon$ and let $X$ be the number of elements is the selected set. Find the distribution of $X$ & $E(X)$.

• Solution:

• $^n$ (i) $x$ can take values $1,3, \ldots, 2 m+1$

• $P(x=2 i+1)=\frac{\left(\begin{array}{c} 2 m+1 \\ 2 i+1 \end{array}\right)}{2^{2 m}}, i=0,1, \cdots m$

• $\begin{aligned} & E(X)=\sum_{i=0}^m(2 i+1) P(X=2 i+1) \\ & =\sum_{i=0}^m(2 i+1) \cdot \frac{\left(\begin{array}{l}2 m+1 \\ 2 i+1\end{array}\right)}{2^{2 m} } \\ & =\sum_{i=0}^m \frac{(2 i+1) \cdot(2 m+1) !}{2^{2 m} (2 i+1) !(2 m-2 i) !} \\ & =\frac{(2 m+1)}{2^{2 m}} \sum_{i=0}^m \frac{2 m !}{2 i !(2 m-2 i) !}=T\end{aligned}$

• $\begin{aligned} & \frac{(2 m+1)}{2^{2 m}} \cdot \sum_{i=0}^m\left(\begin{array}{l} 2 m \\ 2 i \end{array}\right)=\frac{(2 m+1)}{2^{2 m}} \cdot 2^{2 m-1} \\ & =\frac{2 m+1}{2}=\left(m+\frac{1}{2}\right) \end{aligned}$

• A package of 4 IC's contains one defective. $I C$ 's are tested one by one without replacement until the defective is found out. Let $X$ be the number testings required. Find solution of $X$ and $E(X)$.
• the number testings required and $E(X)$.

• The possible values of $X$ are $1,2,3$.

• $P(X=1)=\frac{1}{4}, \quad P(x=2)=\frac{3}{4} \cdot \frac{1}{3}=\frac{1}{4}$

• $P(X=3)=\frac{1}{2}=\left(\frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2}+\frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2}=\frac{1}{2}\right)$

• $E(X)=1 \cdot \frac{1}{4}+2 \cdot \frac{1}{4}+3 \cdot \frac{1}{2}=\frac{9}{4} .$

• Let $X$ be a discrete $r . v$. with disf"

• $P(X=k)=\frac{c}{2^k}, k=0,1, \cdots, n-1, n \geqslant 1$

• Find $c, E(X)$

• Solution:

• $\sum_{k=0}^{n-1} P(x=k)=1$

• $\Rightarrow \sum_{k=0}^{n-1} \frac{1}{k}=1 \Rightarrow c \cdot\left(1+\frac{1}{2}+\cdots+\frac{1}{2^{n-1}}\right)=1$

• $\Rightarrow c \cdot \frac{\left(1-\frac{1}{2^n}\right)}{1-\frac{1}{2}}=\frac{c \cdot\left(2^n-1\right)}{2^{n-1}}=1$

• $\Rightarrow c=\frac{2^{n-1}}{2^n-1}$

• $E(X) =\sum_{k=0}^{n-1} k P(x=k)$

• $=c \sum_{k=0}^{n-1} \frac{k}{2^k}$

• $=c\left[{\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{(n-1)}{2^{n-1}}}\right]$

• $S =\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{(n-1)}{2^{n-1}} .....(1)$

• $\frac{S}{2}=\frac{1}{2^2}+\frac{2}{2^3}+\cdots+\frac{(n-2)}{2^{n-1}}+\frac{(n-1)}{2^n}...(2)$

• Subtract (1) -(2)

• $\frac{S}{2}$

• $=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}-\frac{(n-1)}{2^n}$

• $=\frac{1}{2}\left[\frac{1-\frac{1}{2^{n-1}}}{1-\frac{1}{2}}\right]-\frac{n-1}{2^n}$

• $\Rightarrow S =\frac{2^n-2 n}{2^{n-1}}$

• $E(X) =\frac{2\left(2^{n-1}-n\right)}{2^n-1 .}$

• Let $X$ be a discrete r.u. with probability distribution
• $\begin{array}{lllllllll} x_i: & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ p\left(x_i\right): & 2 k^2 & k & 2 k & 3 k & 2 k & k & 7 k^2 & k^2 \end{array}$
• where $k$ is such that it is a proper prob. dist.
• Find $k, E(X), V(X)$

• $\begin{aligned} & \sum p\left(x_i\right)=1 \\ & \Rightarrow 10 k^2+9 k=1 \Rightarrow 10 k^2+9 k-1=0 \\ & \Rightarrow(10 k-1)(k+1)=0 \Rightarrow k=\frac{1}{10},-1 \end{aligned}$

• $k=-1$ is not possible.

• So the correct value is $k=\frac{1}{10}$

• Then the prob. dist" of $X$ is

• $P(X=-3)=\frac{1}{50}, P(X=-2)=\frac{1}{10}, P(X=-1)=\frac{1}{5}$

• $P(X=0)= \frac{3}{10}, P(X=1)=\frac{1}{5}, P(X=2)=\frac{1}{10},$

• $P(X=3)=\frac{7}{100}, P(X=4)=\frac{1}{100}$

• $E(X)=-3 \times \frac{1}{50}-2 \times \frac{1}{10}-1 \times \frac{1}{5}+0 \cdot \frac{3}{10}+1 \cdot \frac{1}{5}$

• $+2 \cdot \frac{1}{10}+3 \times \frac{7}{100}+4 \times \frac{1}{100}=\frac{19}{100}$

• $E\left(X^2\right)=(-3)^2 \times \frac{1}{50}+(-2)^2 \times \frac{1}{10}+(-1)^2 \times \frac{1}{5}+0^2 \times \frac{3}{10}$

• $+1^2 \cdot \frac{1}{5}+2^2 \cdot \frac{1}{10}+3^2 \times \frac{7}{10}+4^2 \times \frac{1}{100}=\frac{47}{20}$

• $V(X)=E\left(X^2\right)-{E(X)}^2 \cong 2.3139$

• In this distn find $P(|X| \geqslant 2)$

• $P(|X| \geqslant y=P(X \geqslant 2)+P(X \leqslant-2)$

• $=P(X=2)+P(X=3)+P(X=4)+P(X=-3)+P(X=-2)$

• $=\ldots . .$

• Let $X$ be a discrete random variable with possible values $-2,-1,122$. It is given that $p(X=-2)=\frac{1}{3}, p(X=2)=\frac{13}{60}$ but probabilities of -122 are not given. Further it is known that $E(X)=-\frac{17}{60}$ determine.

• Solution:

• $P(X=-2)+P(X=2)+P(X=-1)+P(X=1)=1$

• $\Rightarrow \frac{1}{3}+\frac{13}{60}+q+p=1 \Rightarrow p+q=\frac{q}{20} \cdots(1)$

• $E(X)=-2 \times \frac{1}{3}+2 \times \frac{13}{60}+q(-1)+p(1)=-\frac{17}{60}$

• $\Rightarrow p-q=-\frac{1}{20} \ldots(2)$

• Solving (1) & (2), we get $p=\frac{1}{5} 2 q=\frac{1}{4}$.

• So $P(X=1)=\frac{1}{5}, \quad P(X=-1)=\frac{1}{4}$.

• Suppose $X$ is a discrete random variable with distribution given by
• $P(X=-1)=\frac{1-2 \alpha}{3}, \quad P(X=1)=\frac{1+2 \alpha}{3}, P(X=0)=\frac{1}{3}$
• where $\alpha$ is a real number. Find range of $\alpha$. Also determine values of $\alpha$ for which $\operatorname{Var}(X)$ is maximum or minimum.

• $\frac{1-2 \alpha}{3}+\frac{1+2 \alpha}{3}+\frac{1}{3}=1$

• $0 \leq \frac{1-2 \alpha}{3} \leq 1 \Rightarrow 0 \leq 1-2 \alpha \leq 3$

• $0 \leq \frac{1+2 \alpha}{3} \leq 1 \Rightarrow 0 \leq 1+2 \alpha \leq 3 \Rightarrow-\frac{1}{2} \leq(1)$

• Taking intersection of the regions in (1) and(2), we get the range of $\alpha$ as $-\frac{1}{2} \leqslant \alpha \leqslant \frac{1}{2}$.

• $E(X) =(-1) \times \frac{1-2 \alpha}{3}+(1) \times \frac{1+2 \alpha}{3}+0 \cdot \frac{1}{3} \quad =\frac{4 \alpha}{3}$

• $E\left(X^2\right) =(-1)^2 \times \frac{1-2 \alpha}{3}+(1)^2 \times \frac{1+2 \alpha}{3}+0^2 \cdot \frac{1}{3} \quad =\frac{2}{3} \cdot$

• $\operatorname{Var}(X) =E\left(X^2\right)-{E(X)}^2=\frac{2}{3}-\frac{16 \alpha^2}{9}$

• $\operatorname{Var}(X)$ is maXimum when $\alpha=0$

• So, The minimum value of $\operatorname{Var}(X)$

• $\operatorname{Var}(X)=E\left(X^{\prime}\right)$

• $\operatorname{Var}(X)$ is maximum

• when $\alpha=0$

• So, The minimum value of $\operatorname{Var}(X)$ :

• $g(\alpha)=\frac{2}{3}-\frac{16 \alpha^2}{9}$

• $g^{\prime}(\alpha)=-\frac{32 \alpha}{9}$

• $>0 \text { if } -\frac{1}{2} \leq \alpha<0$

• $<0 \text { if } ~ 0<\alpha>\frac{1}{2}$

• $g^{\prime}(\alpha)=-\frac{32 \alpha}{9} >0 \text { of }$

• $-\frac{1}{2} \leqslant \alpha<0 \text { of } ~$

• $0<\alpha \leqslant \frac{1}{2}$

• So the minimum value of $g(\alpha)$ is attained at $\alpha=-\frac{1}{2} , \alpha=\frac{1}{2}$.

• So the minimum value of $g(\alpha)$ is attained at $\alpha=-\frac{1}{2} , \alpha=\frac{1}{2}$.

• Independent questions are posed in a quiz to a candidate. If the candidate fails to answer he/she has to leave the quiz.
• Let the probe of answering a question be $p$.

• It is known that the candidate answers an even number of questions 2 then fails is 0.9 . what is $p$ ?

• solution:

• $X \rightarrow$ the number of questions asked to the candidate.

• $P(X=k)=p^{k-1}(1-p), \quad k=1,2, \ldots$

• $1-p=q$

• Now it is given that the candidate answer an even number of questions correctly of 0.9 .

• $\Rightarrow \sum_{k=0}^{\infty} p(X=2 k+1)=\sum_{k=0}^{\infty} p^{2 k} q=\frac{q}{1-p^2}=0.9$

• $\Rightarrow\frac{1-p}{(1-p)(1+p)}=0.9=1+p=\frac{10}{9}$

• $\Rightarrow p=1 / 9 .$

• A missile can successfully hit a target with prob 0.75 . If three successful hits can destroy the largest completely, how many missiles must be fired simultaneously so that the prob.of completely destroying the target is not less than 0.95 ?

• Solution:

• Let $n$ missile be fired simultaneously & $X \rightarrow$ no. of missiles hitting the target.

• $X \sim \operatorname{Bin}(n, 0.75), \quad p=0.75$

• We want $P(X \geqslant 3) \geqslant 0.95$

$\begin{aligned} & \Rightarrow \quad P(X<3) \leqslant 0.05 \\ & \Rightarrow p(X=0)+p(X=1)+p(X=2) \leqslant 0.05 \\ & \Rightarrow\left(1-\frac{3}{4}\right)^n+\left(\begin{array}{l} n \\ 1 \end{array}\right)\left(1-\frac{3}{4}\right)^{n-1}\left(\frac{3}{4}\right) +\left(\begin{array}{l} n \\ 2 \end{array}\right)\left(1-\frac{3}{4}\right)^{n-2} \cdot\left(\frac{3}{4}\right)^2 ~ \leq 0.05 \\ & =\left(\frac{1}{4}\right)^n+n \cdot\left(\frac{1}{4}\right)^{n-1} \cdot \frac{3}{4} +\frac{n(n-1)}{2} \cdot\left(\frac{1}{4}\right)^{n-2} \cdot\left(\frac{3}{4}\right)^2 \leq 0.0 \\ & \end{aligned}$

• After simplification this term reduces to

• $10\left(9 n^2-3 n+2\right) \leq 4^n$

• This condition is satisfied for $n \geqslant 6$.

• So, minimum number of firings must be 6 .

• An Item is defective with prob. (0.01) What is the prob that in a pack of ten items none or only ore defective is there?

• Solution:

• $X \rightarrow$ no. of defectives out of 10

• $X \sim \operatorname{Bin}(10,0.01)$.

• $\begin{aligned} p(X \leq 1) & =p(X=0)+p(X=1) \\ & =(0.99)^{10}+10(0.92)^9(0.01) \cong 0.9957 \end{aligned}$