mathematics-class-12-unit-13-chapter-07-probability-conditional-probability-independence-l-7-10-dd2j3312f0k

### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> Conditional probability </primary>
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- Let $A$ and $B$ be any two events with $P(B)>0$. Then the conditional probability of $A$ given that event $B$ has already occurred
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<footer style = "font-size: 18px"> $\rightarrow$ Probability lecture-7 $\rightarrow$ <span style = "color:blue"> Conditional probability </span> $\rightarrow$ Conditional probability $\rightarrow$ General multiplication rule </footer>

### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> Conditional probability </primary>
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- $ P(A \mid B)=\frac{P(A \cap B)}{P(B)} ......(1) $

- From relation (1) we can write
- $\Rightarrow P(A \cap B)=P(B) P(A \mid B)....(2)$

- Similarly if $P(A)>0$, we can define

- $
\begin{aligned}
& P(B \mid A)= \frac{P(A \cap B)}{P(A)} \\\\
& \Rightarrow P(A \cap B)=P(A) P(B \mid A)....(3)
\end{aligned}
 $

- (2) and (3) are called multiplication rules.
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<footer style = "font-size: 18px">Probability lecture-7 $\rightarrow$ Conditional probability $\rightarrow$ <span style = "color:blue"> Conditional probability </span> $\rightarrow$ General multiplication rule $\rightarrow$ Proof </footer>

### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> General multiplication rule </primary>
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- Let $A_1, A_2, \ldots A_n$ be events with $P\left(\bigcap_{i=1}^n A_i\right)>0$
- (This will ensure that all conditional probabilities are well defined).

- Then,

- $P\left(\cap_{i=1}^n A_i\right)=P\left(A_1\right) P\left(A_2 \mid A_1\right) P\left(A_3 \mid A_1 \cap A_2\right)$

- $\cdots P\left(A_n \mid n_{i=1}^{n-1} A_i\right)$....(4)

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![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/710c5cff-14ad-4b41-d018-ed0553797300/public)

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<footer style = "font-size: 18px">Conditional probability $\rightarrow$ Conditional probability $\rightarrow$ <span style = "color:blue"> General multiplication rule </span> $\rightarrow$ Proof $\rightarrow$ Proof </footer>

### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> Proof </primary>
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- The relation (4) can be proved using the Principle of Mathematical Induction.

- For $n=1$, the statement is always true as it reduces to $P\left(A_1\right)=P\left(A_1\right)$
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<footer style = "font-size: 18px">Conditional probability $\rightarrow$ General multiplication rule $\rightarrow$ <span style = "color:blue"> Proof </span> $\rightarrow$ Proof $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> Proof </primary>
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- Since $B_1, B_2, \ldots, B_n$ are pairwise disjoint, the sets (events)
- $A \cap B_1, A \cap B_2, \ldots, A \cap B_n$ will also be pairwise disjoint

- $A \cap B_1 \subset B_1, 
- A \cap B_2 \subset B_2, ...,$

- $A \cap B_n \subset B_n$

- Then by the axiom of additivity, we get

- $
 P(A)=P\left(\bigcup_{i=1}^n\left(A \cap B_i\right)\right)=\sum_{i=1}^n P\left(A \cap B_i\right)
 $

- $=\sum_{i=1}^n P\left(A \mid B_i\right) P\left(B_i\right)$
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<footer style = "font-size: 18px">Theorem of total probability $\rightarrow$ Proof $\rightarrow$ <span style = "color:blue"> Proof </span> $\rightarrow$ Problem-1 $\rightarrow$ Bayes theorem </footer>

### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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-  A fair coin is tossed once. If a head comes up, a fair die is tossed once, and of a tail comes up, a fair die is tossed tine. We want to find the probability that at least a six is observed.

- Solution:  $\mathrm{H} \rightarrow$ head comes up
- $T \rightarrow$ tail comes up
- $A \rightarrow a$ six is observed.

- $\begin{aligned} P(A) & =P(A \mid H) P(H)+P(A \mid T) P(T) .\end{aligned}$

- $\begin{aligned}
& =\frac{1}{6} \cdot \frac{1}{2}+\left(1-\left(\frac{5}{6}\right)^2\right) \times \frac{1}{2}=\frac{17}{72}
\end{aligned}$
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### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> Proof </primary>
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- $
\begin{aligned}
& P\left(B_r \mid A\right)=\frac{P\left(B_r \cap A\right)}{P(A)} \\\\
&= \frac{P\left(A \mid B_r\right) P\left(B_r\right)}{\sum_{i=1}^n P\left(A \mid B_i\right) P\left(B_i\right)} \end{aligned}$

- $= \frac{\rightarrow \text { using multiplication } }{\rightarrow \text {using theorem of total probability}}$

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### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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-  A computer manufacture process chips from 3 supplies $B_1, B_2, B_3$ in proportion $\frac{2}{5}, \frac{3}{10}$ and $\frac{3}{10}$ respectively. It is known from experience that 1 % of chips from $B$, are defective. 5% from $B_2$ are defective and 10 % from $B_3$ are defective. A clip is randomly selected from the collection of the manufacturer. It is found to be defective. What is the probability that it was supplied by $B$,.
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### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $A \rightarrow$ chip is defective

- $P(A)=\sum_{i=1}^3 P\left(A \mid B_i\right) P\left(B_i\right)$
- $B_i \rightarrow$ the chip is supplied by $B_i$

- $\begin{aligned} > & =\frac{1}{100} \times \frac{2}{5}+\frac{5}{100} \times \frac{3}{10}+\frac{10}{100} \times \frac{3}{10}=0.049 \end{aligned}$

- $\begin{aligned}P\left(B_1 \mid A\right) & =\frac{P\left(A \mid B_1\right) P\left(B_1\right)}{P(A)}=\frac{\frac{1}{100} \times \frac{2}{5}}{0.049}=\frac{4}{49} \end{aligned}$

- $\begin{aligned}  P\left(B_2 \mid A\right) & =\frac{P\left(A \mid B_2\right) P\left(B_2\right)}{P(A)}=\frac{\frac{5}{100} \times \frac{3}{10}}{0.049}=\frac{15}{49} \end{aligned}$

- $\begin{aligned} P\left(B_3 \mid A\right) & =\frac{P\left(A \mid B_3\right) P\left(B_3\right)}{P(A)}=\frac{\frac{10}{100} \times \frac{3}{10}}{0.049}=\frac{30}{49} .\end{aligned}$

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### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Independence of Events

- if occurrence of $B$ does not affect the prob of $A$, then $P(A \mid B)=P(A)$
- $
\begin{aligned}
& \Rightarrow \frac{P(A \cap B)}{P(B)}=P(A) \\
& \Rightarrow P(A \cap B)=P(A) P(B) \rightarrow(1)
\end{aligned}
 $

- If occurrence of $A$ does not affect the prob B
- then $P(B \mid A)=P(B)$
- $
\Rightarrow \quad \frac{P(B \cap A)}{P(A)}=P(B) \Rightarrow P(B \cap A)=P(B) P(A)
 $

- We define events $A$ and $B$ to be independent if $P(A \cap B)=P(A) P(B)$
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### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- Suppose two fair dice are tossed simultaneously
- Let $A \rightarrow$ even number on the first die
- $B \rightarrow$ even number on the second die
- $
\begin{aligned}
& P(A)=\frac{18}{36}=\frac{1}{2}, P(B)=\frac{18}{36}=\frac{1}{2} \\
& P(A \cap B)=\frac{9}{36}=\frac{1}{4}
\end{aligned}
 $

- So $P(A \cap B)=P(A) P(B)$.

- So events $A$ and $B$ are independent

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| Column 1 | Column 2 | Column 3 | Column 4 |
| -------- | -------- | -------- | -------- |
| 1 ,1   | 1 ,2     | .....   | 1 ,6   |
| 2 ,1    | 2 ,2     | ....  | 2 ,6   |
| ....    | ....     | .... | ....    |
| 6 ,1    | 6 ,2     | .... | 6 ,6    |

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

### <Success style="font-size:25px"> Probability Conditional Probability Independence L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- In the previous example, let $C \rightarrow$ sum is even.
- $
\begin{aligned}
& P(C)=\frac{18}{36}=\frac{1}{2}, P(A \cap C)=\frac{9}{36}=\frac{1}{4} \\\\& P(B \cap C)\end{aligned}
 $

- $\begin{aligned}=\frac{9}{36}=\frac{1}{4} \text {. So } A, B, C \text { are pairwise independence } \end{aligned} $

- $\begin{aligned} & P(A \cap B \cap C)=\frac{1}{4} \neq P(A) P(B) P(C)\end{aligned} $

- So $A, B, C$ are not mutually independent.
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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>