mathematics-class-12-unit-13-chapter-08-probability-some-problems-on-discrete-distributions-l-8-10-f5px6hjxy64

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- 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
 $
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<footer style = "font-size: 18px"> $\rightarrow$ Probability lecture-8 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- 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}$
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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 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 .}$
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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ 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 . .$

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<footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-5 $\rightarrow$ Problem-6 </footer>

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
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-  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 $P(X=1)=\frac{1}{5}, \quad P(X=-1)=\frac{1}{4}$.
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Problem-6 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $
\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 
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<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 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 $\alpha=-\frac{1}{2} , \alpha=\frac{1}{2}$.

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<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-7 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Problem-7 </primary>
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-  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 $p$.
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-7 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 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} $
- $\Rightarrow p=1 / 9 .$
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<footer style = "font-size: 18px">Problem-7 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-8 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Solution </primary>
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$
\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}
$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-8 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-9 </footer>

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 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 .
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<footer style = "font-size: 18px">Problem-8 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-9 $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Probability Some Problems On Discrete Distributions L-8 </Success>
### <primary style="font-size:24px"> Problem-9 </primary>
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-  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}
 $
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-9 </span> $\rightarrow$ Thank you $\rightarrow$  </footer>