mathematics-class-12-unit-13-chapter-06-probability-some-more-problems-on-probability-l-6-10-mqudvo9iseo

### <Success style="font-size:25px"> Probability Some More Problems On Probability L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- So, $ ~ $ $P\left(A_1\right)=\frac{6}{11}, \quad P\left(A_2\right)=\frac{3}{11}, P\left(A_3\right)=\frac{2}{11}$. 
- $P\left(E \mid A_1\right)=\frac{1}{6}, \quad P\left(B E \mid A_2\right)=\frac{1}{5}, P\left(E \mid A_3\right)=\frac{1}{4}$

- By theorem of total probability
- $ P(E)=\sum_{i=1}^3 P\left(E \mid A_i\right) P\left(A_i\right) $
- $ =\frac{1}{6} \cdot \frac{6}{11}+\frac{1}{5} \cdot \frac{3}{11}+\frac{1}{4} \cdot \frac{2}{11}=\frac{21}{110} $

- $P\left(A_2 \mid E\right)=\frac{P\left(E \mid A_2\right) P\left(A_2\right)}{P(E)}=\frac{\frac{1}{5} \cdot \frac{3}{11}}{\frac{21}{110}}=\frac{2}{7}$

- (by Bayes' Theorem).
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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability Some More Problems On Probability L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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-  Let $B_i$ be the event that $S_i$ hits the target., $i=1,2,3$.
- Then $B_1, B_2, B_3$ are independent.
- Then $P\left(B_1\right)=\frac{2}{5}, P\left(B_2\right)=\frac{1}{3}, P\left(B_3\right)=\frac{3}{7}$.
- Let $A$ be the event that these are two hits.

- Then  $\quad A=\frac{\left(B_1 \cup \cap B_2 \cap B_3^c\right) \cup\left(B_1 \cap B_2^c \cap B_3\right)}{\cup\left(B_1^c \cap B_2 \cap B_3\right)} $

- So,   P(A)=P $ \left(B_1 \cap B_2 \cap B_3^c\right)+P\left(B_1 \cap B_2^c \cap B_3\right) $
- $ +P\left(B_1^C \cap B_2 \cap B_3\right) $

- $ =P\left(B_1\right) P\left(B_2\right) P\left(B_3^c\right)+P\left(B_1\right) P\left(B_2^c\right) P\left(B_3\right) $
- $ +P\left(B_1^c\right) P\left(B_2\right) P\left(B_3\right) $

- $ =\frac{2}{5} \cdot \frac{1}{3} \cdot \frac{4}{7}+\frac{2}{5} \cdot \frac{2}{3} \cdot \frac{3}{7}+\frac{3}{5} \cdot \frac{1}{3} \cdot \frac{3}{7}=\frac{29}{105} $

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

### <Success style="font-size:25px"> Probability Some More Problems On Probability L-6 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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-  A computer producing factory has only two plants say $T_1$ and $T_2$. Plant $T_1$ produces 20% and

- plant $T_2$ produces 80% of the total computers produced. 7% of the computers produced by the factory turn out to be defective. Further it is
- P computer is defective of it is produced is $T_1$ ) =10 P (computer is defective | it is produced in $T_2$ ) A computer produced in the factory is randomly, selected and it is not defective. What is the probability that it is produced is $T_2$ ?

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Probability Some More Problems On Probability L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Let $A$ be the event that the computes is defective. $B_i \rightarrow $ it is produced in $T_i, i=1,2$,

- Then $P\left(B_1\right)=\frac{1}{5}, P\left(B_2\right)=\frac{4}{5}, P(A)=\frac{7}{100}$

- Now $P(A)=P\left(A \mid B_1\right)$

- P$ \left(B_1\right)+P\left(A \mid B_2\right) P\left(B_2\right)$ (by Theorem if Total Probability)

- Let $P\left(A \mid B_2\right)=\alpha$. Then $P\left(A \mid B_1\right)=10 \alpha$ Then from equation (1) :

- $ \frac{7}{100}=10 \alpha \cdot \frac{1}{5}+\alpha \cdot \frac{4}{5}=\frac{14 \alpha}{5}$

- $ \Rightarrow \quad \alpha=\frac{1}{40} \text { So, } ~ P\left(A \mid B_2\right)=\frac{1}{40} $
\&  $ P\left(A \mid B_1\right)=\frac{1}{4} $

- Required probability $=P\left(B_2 \mid A^c\right)=\frac{P\left(B_2 \cap A^c\right)}{P\left(A^c\right)} $

- $ =\frac{P(A^c \mid B_2) P(B_2)}{P(A^c)}=\frac(1-\frac{1}{40}) \times $
- $ \frac{4}{5}{\frac{93}{100}}=\frac{78}{93}$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability Some More Problems On Probability L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Total number if ways of choosing (drawing)
- $x_1, x_2 x_3$ is $\quad 3\times 5 \times 7=105$
- (i) $x_1$ can take 3 values If $x_1=1$ or 3, then $x_2+x_3$ 
- must be even So either $x_2$ \& $x_3$ $ax ~ $ even $\rightarrow 2 \times 3=6$ 
- cases So either $x_n$ \& $x_3$ are odd $\rightarrow 3 \times 4=12$ cases. 
- So total cases $=2 \times(6+12)=18 \times 2=36$ cases If $x_1=2$, then $x_2+x_3$ must be odd.
- $x_2$ odd \& $x_3$ are $\rightarrow 3 \times 3 \rightarrow 9$ cases
- $x_2$ even \& $x_3$ are $\rightarrow 2 \times 4 \rightarrow 8$ cases
- So in this cases,  total number of cases $=9+8=17$.

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

### <Success style="font-size:25px"> Probability Some More Problems On Probability L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- So total no. of cases in which $x_1+x_2+x_3$ is
- odd =36+17=53
- required probability = $\frac{53}{105}$
- (iii) Let d be the common difference

|D               | -1    | 0   | 1  | 2 |   3   |
| --------       | ----- | ----| ---| --|       |
| No of cases    | 1     | 3   | 3  | 3 |      1|
- required Probabilty = $\frac{11}{105}$

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