mathematics-class-12-unit-13-chapter-15-probability-l-5-6-5ds1tllqe4g

### <Success style="font-size:25px"> Probability L-5 </Success>
### <primary style="font-size:24px"> Unconditional probability </primary>
<hr>
<main>

- What in the unconditional probability that the $2^{\text {rd }}$ ball drawn is Green?
- $
\text { i.e } P\left(G_2\right)
 $

- Now the event $G_2=R_1 G_2 \quad G_1 G_2$
- $
 P\left(G_2\right)=P\left(R_1 G_2 \cup G_1 G_2\right)
 $

- $\therefore P\left(G_2\right)  =P\left(R_1 \cap G_2\right)+P\left(G_1 \cap G_2\right) =P\left(G_2 \mid R_1\right) \cdot P\left(R_1\right)+P\left(G_2 \mid G_1\right) \cdot P\left(G_1\right)$
 
- $\left.=\frac{8}{19} \times \frac{3}{5}+\frac{7}{19} \times \frac{2}{5}=\frac{24+14}{19 \times 5}=\frac{38}{19 \times 5}=\frac{2}{5}\right)$

- Verify that the uncoditional probabilities of $R_2=\frac{3}{5}$
</div>
<div style='float: right; width:1%;'>

</div>
<div style='float: right; width:35%;'>

</main>
<hr>
<footer style = "font-size: 18px">Probability lecture-15 $\rightarrow$ Conditional probabiity $\rightarrow$ <span style = "color:blue"> Unconditional probability </span> $\rightarrow$ Unconditional probabiity $\rightarrow$ Unconditional probabiity </footer>

### <Success style="font-size:25px"> Probability L-5 </Success>
### <primary style="font-size:24px"> Unconditional probabiity </primary>
<hr>
<main>

- $P\left(R_3\right)=\frac{3}{5} \times \frac{11}{19} \times \frac{10}{18}+\frac{3}{5} \times \frac{8}{19} \times \frac{4}{18}+\frac{2}{5}$

- $\times \frac{2}{19} \times \frac{11}{18}+\frac{2}{5} \times \frac{7}{19} \times \frac{12}{18} $

- $ =\frac{1}{5 \times 19 \times 18}(330+264+264+168)$

- $=\frac{102657}{5 \times 19 \times 18}=\frac{3}{5}$
</div>
<div style='float: right; width:1%;'>

</div>
<div style='float: right; width:34%;'>

</main>
<hr>
<footer style = "font-size: 18px">Unconditional probability $\rightarrow$ Unconditional probabiity $\rightarrow$ <span style = "color:blue"> Unconditional probabiity </span> $\rightarrow$ Problem 1 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- (a) $P(5)=P(5 / \text { fair die }) \times P(\text { fair die }) $

- $+P(5 / \text { fake die }) \times P(\text { fake die}) $

- $=\frac{1}{6} \times \frac{3}{4}+\frac{2}{3} \times \frac{1}{4} $

- $=\frac{3}{24}+\frac{4}{24}=\frac{7}{24}   $

- (b) $P(\text { chosen the fake die } \mid \text { result is 5) } \longrightarrow$ bayes' theorem

</div>

![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/8939a23f-dcec-4714-4615-4537a38c6400/public)

</main>
<hr>
<footer style = "font-size: 18px">Unconditional probabiity $\rightarrow$ Problem 1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Bayes' theorem $\rightarrow$ Bayes' theorem </footer>

### <Success style="font-size:25px"> Probability L-5 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
<hr>
<main>

- Suppose you have 10 coins numbered $1,2,3 \cdots 10$. Suppose Prob of getting  it in single toss of the $i^{\text {th }}$ coin in $\frac{i}{10}, i=1,2, \ldots 10$.
- You choose a coin randomly and toss it. If the result you on got in $ H$, what in the probability that you have chosen the $5^{\text {th }}$ coin.?

</div>
<div style='float: right; width:1%;'>

</div>

</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Bayes' theorem $\rightarrow$ Bayes' theorem $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-5 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
<hr>
<main>

- Suppose a student is answering a question an MCQ question with 5 optionsof which only one is correct.

- Now let P be the probability that the student gives the answer i.e. he randomly ticks and let 1-p be the probability that he knows the answer and therefore he ticks correctly.

- Suppose the student ticks correctly. What is the probability that he gives the answer.

- We want to compute P(guess/ ticked rightly)

</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<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 L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- Let $E$ be the event of selecting bag $A$.            
- $B$ be the event of getting $1W+1R$.

- We want to compute $P(E \mid B)$

- Using Bayers' Theorem: $P(E \mid B)=\frac{P(B \mid E) \cdot P(E)}{P(B)}$

- Now $P(E)=\frac{1}{3}$ and $P(B \mid E)=\frac{3}{{}^6{c_2}}=\frac{3}{\frac{6!}{2!4!}}$

- $=\frac{3}{\frac{5 \times 6}{2}}=\frac{1}{5} .$

- Similary ,$P(1W+1 R \mid Bag B)^2=\frac{2}{{}^4c_2}=\frac{2}{\frac{4 !}{2 ! 2 !}}=\frac{2}{6}=\frac{1}{3} $

- $P(1W+1 R \mid Bag C)=\frac{12}{12 c_2}=\frac{12}{\frac{11\times12}{2}}=\frac{2}{11}$

</div>

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

### <Success style="font-size:25px"> Probability L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $P\left(K O L K A T A \mid AT\right) =\frac{P\left(AT \mid \text { KOLKATA }\right) \times P(K O L K A T A)}{P\left(AT\right)}$

- Now

- $P(A T \mid \text { KOLKATA) } : \frac{1}{6}$

- $P(A T \mid \text { KATHMANDU)} : \frac{1}{8} $

- $P(A T \mid \text { MUSCAT) }=\frac{1}{5}$

- $P(K O L K A T A \mid A T) =\frac{\frac{1}{6} \times \frac{1}{3}}{\frac{1}{6} \times \frac{1}{3}+\frac{1}{8} \times \frac{1}{3}+\frac{1}{5} \times \frac{1}{3}}$

- $=\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{8}+\frac{1}{5}} =\frac{\frac{1}{6}}{\frac{40+30+48}{5 \times 6 \times 8}} $

- $=\frac{40}{40+30+48}=\frac{40}{118}=\frac{20}{59}$

</div>

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

### <Success style="font-size:25px"> Probability L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- Can it be 1? Because since viral won all the previous tosses, therefore he will win the fifth toss.                         
- Can it be $\frac{1}{5}$ ? Because Virat won all the 4 tosses therefore now the Australian captain will have $\frac{4}{5}$ probability of winning.                         
- Can it be 0 ? Because now Australian captain $X$ will have to win the toss                         
- $\therefore ~ P \text { (virat winning) }=0$            
- Correct option is (D) i.e. Answer is $\frac{1}{2}$

- $P$ (virat winning $5^{\text {th }}$ toss $\mid$ he won all the previus tosses $)$ is same as $P$ (virat winning the toss) as there are independent events.
 
</div>

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