mathematics-class-12-unit-13-chapter-05-probability-further-problems-on-probability-l-5-10-izuzkiwi5wy

### <Success style="font-size:25px"> Probability Further Problems On Probability L-5 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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- Let $E$ and $F$ be independent and $P(E)+P(F)=1$

- Also let $P(E \cap F)=\frac{2}{9}$ and $P(E)>P(F)$. Find $P(E)$.

- <ins>**Solution:-**</ins>    
- Let $P(E)=p, \Rightarrow P(F)=1-p$
- $
\begin{aligned}
& P(E \cap F)=P(E) P(F)=P(1-p)=\frac{2}{9} . \\\\
& \Rightarrow P \text { is } \frac{1}{3} \text { or } \frac{2}{3}
\end{aligned}
 $

- $\Rightarrow p$ is $\frac{1}{3}$ or $\frac{2}{3}$

- So $P(E)=\frac{2}{3}$

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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability Further Problems On Probability L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Again from (1) $\Rightarrow P(E)-P(E \cap F)<P(E)-P(E) P(F) $     
- $ \Rightarrow P(E \cap F^c)<P(E)(1-P(F)) $     
- $ \Rightarrow P(E \cap F^c)<P(E) P(F^c) $     
- $ \Rightarrow \frac{P(E \cap F^c)}{P(F^c)}<P(E) $     
- $ \Rightarrow P(E \mid F^c)<P(E) $

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

### <Success style="font-size:25px"> Probability Further Problems On Probability L-5 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- Two points are chosen at random on a line segment of length $9 \mathrm{~cm}$. Find the probability that the distance between these two points is less than 3 cm

- Solution

- $
\begin{aligned}
& x-y<3 \\
& y-x<3
\end{aligned} $

- We can consider the point $(x, y)$ in a square in a two dimensional plane.
- $
\begin{aligned}
\text { The read prob. } & =\frac{\text { area } of \text { the shaded region }}{\text { Total area }} \\\\ =\frac{81-36}{81} & =\frac{5}{9}
\end{aligned}
 $
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![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/efab468e-e514-43d3-b884-8ffdf0ee9d00/public)

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

### <Success style="font-size:25px"> Probability Further Problems On Probability L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Let $A$ denote the event that ball from $U_2$ is white. Let $B_1$ be the event that these is head on the crim and $B_2$ be the event that there is tail on the coin. $P\left(B_1\right)=P\left(B_2\right)=\frac{1}{2}$.
- $\left(A \mid B_1\right)=P\left(\right.$ white ball is drawn from $U_2 \mid$ white  bell in drawer $P$ (white ball is drawn from $U_1$ )
- $+P$ (white ball is drawn from $\left.U_2\right)$ red ball is drawn $P$ ( red ball is drown from $\left.U_1\right)$

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

### <Success style="font-size:25px"> Probability Further Problems On Probability L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ =1 \cdot \times \frac{3}{5}+\frac{1}{2} \times \frac{2}{5}=\frac{4}{5}$
- $P\left(A \mid B_2\right)=P$ (white ball is drown from $U_2$ | two whites balls are drawn from $\left.U_1\right) P\left(\right.$ tow whites balliare drain from $\left.U_1\right)$
- $+P$ (white ball is drawn from $U_2$ | one white 2 one red ball is dream from $\left.U_1\right) P$ (one white and one red ball is drawn from $\left.U_1\right)+P$ (white ball is drawn from $U_2$ ) two red balls are drawn from $\left.U_{1}\right) P$ (for red balls are drawn from $U_1$ ).

- $=1 \cdot \frac{\left(\begin{array}{l}^3 C_ 2\end{array}\right)}{\left(\begin{array}{l} ^5 C_ 2\end{array}\right)}+\frac{2}{3} \times \frac{4\left(\begin{array}{l} ^3 C_ 1\end{array}\right)\left(\begin{array}{l} ^2 C_ 1\end{array}\right)}{\left( ^5 C_ 2 \right)}+\frac{1}{3} \cdot \frac{\left(\begin{array}{l} ^2 C_ 2\end{array}\right)}{\left(\begin{array}{l} ^5 C_ 2\end{array}\right)}$
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<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-6 </footer>

### <Success style="font-size:25px"> Probability Further Problems On Probability L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Again using the Theorem of total Probability
- $
\begin{aligned}
 P(A) & =P\left(A \mid B_1\right) P\left(B_1\right)+P\left(A \mid B_2\right) P\left(B_2\right) \\\\
& =\frac{4}{5} \cdot \frac{1}{2}+\frac{11}{15} \cdot \frac{1}{2}=\frac{23}{30} .
\end{aligned}
 $

- To find $P(B, \mid A)$ we use Bayes Theorem:
- $
\begin{aligned}
& P\left(B_1 \mid A\right)=\frac{P\left(A \mid B_1\right) P\left(B_1\right)}{P\left(A \mid B_1\right) P\left(B_1\right)+P\left(A \mid B_2\right) P\left(B_2\right)} \\\\
& =\frac{\frac{4}{5} \cdot \frac{1}{2}}{\frac{23}{30}}=\frac{12}{23} .
\end{aligned}
 $

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

### <Success style="font-size:25px"> Probability Further Problems On Probability L-5 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- There are $n$ urns numbered $1,2, \ldots n$ each containing $(n+1)$ balls. Urn contains $i$ white. balls $2(n+1-i)$ red balls, $i=1,2 \ldots, n$. An urn is selected and a ball is drawn from it. Let $U_i$ denote the event that urn $i$ is selected and let $w$ be the event that a white ball is drawn from the selected urn. Further, suppose -that $E$ denotes the event that an even numbered urn is selected.

- (i) Let $p\left(U_i\right) \propto i, \quad i=1, \ldots, n$. Find $\lim _{n \rightarrow \infty} p(w)$

- (ii) If $P\left(U_i\right)=c, i=1 \ldots, n$, where $c$ is a constant, find $p\left(U_n \mid W\right)$.

- (iii) If $p\left(v_i\right)=\frac{1}{n}, i=\ldots, n$ and $n$ is an even positive integer, find $\hat{P}(\omega \mid E)$.
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability Further Problems On Probability L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- The Theorem of total probability gives 
- $ P(W)  =\sum_{i=1}^n p\left(W \mid U_i\right) p\left(U_i\right) $

- $ =\sum_{i=1}^n \frac{i}{n+1} \cdot \frac{2 i}{n(n+1)}=\frac{2}{n(n+1)^2} \sum_{i=1}^n i^2 $

- $ =\frac{2}{n(n+1)^2} \cdot \frac{n(n+1)(2 n+1)}{6}=\frac{2(2 n+1)}{6(n+1)} $

- $\lim _{n \rightarrow \infty} p(w)$ = $\frac{2 \times 2}{6}=\frac{2}{3}$

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

### <Success style="font-size:25px"> Probability Further Problems On Probability L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $
\begin{aligned}
& P(A \cap B)=P(A) P(B) \\\\
\Rightarrow & \frac{n(A \cap B)}{n(S)}=\frac{n(A)}{n(S)} \cdot \frac{n(B)}{n(S)} \\\\
\Rightarrow & n(A \cap B) n(S)=n(A) n(B) \\\\
\Rightarrow \quad & 10 n(A \cap B)=4 n(B)
\end{aligned}
 $                     
- So only possible values of $n(B)$ can be 5 or 10 .         
- If $n(B)=5$, then $n(A \cap B)=2$ and of $n(B)=10$, then $n(A \cap B)=n(A)=4$.

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