mathematics-class-12-unit-13-chapter-14-probability-l-4-6-uxlfp7xuyuk

### <Success style="font-size:25px"> Probability L-4 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- Suppose A and B throw a pair of dice alternatively. The game stops if A throws a total sum of 9 or B throws a total sum of 6.

- Assuming A starts first what is the probability that B finishes the game?

- Solution:

- Let $E_1$ be the event that the sum of frees $=9$

- This can happen with the following cases.

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<footer style = "font-size: 18px"> $\rightarrow$ Probibility lecture-14 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $(3,6)(4,5),(5,4) (6,3)$.

- $
 P\left(E_1\right)=\frac{4}{36}=\frac{1}{9}
 $

- Let $E_2$ be the event that the sum in 6 primitives are $(1,5),(2,4),(3,3),(4,2),(5,1)$

- And $\therefore P\left(\bar{E}_2\right)=\frac{5}{36}$

- We know that A starts the game.

- $\therefore$ A finishes the game if
 A throws 9 first home $\frac{1}{9}$
- A throws other than 9 then $B$ throws other than 6 then if thrown 9
- $
\frac{8}{9} \times \frac{31}{36} \times \frac{1}{9}
 $
- $
\begin{aligned}
& \frac{8}{9} \times \frac{31}{36} \times \frac{8}{9} \times \frac{31}{36} \times \frac{1}{9} \\
 = & \frac{1}{9} \times\left(\frac{8 \times 31}{9 \times 36}\right)^2
\end{aligned}
 $

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<footer style = "font-size: 18px">Probibility lecture-14 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Probability L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Total probabilist that $A$ finishes the game

- $\text { in } \begin{aligned}& \frac{1}{9}+\frac{1}{9} \cdot\left(\frac{8}{9} \times \frac{31}{36}\right)+\frac{1}{9}\left(\frac{8}{9} \times \frac{31}{36}\right)^2+\cdots \cdot \end{aligned}$

- $ \begin{aligned}= & \frac{1}{9}\left(1+\left(\frac{8}{9} \times \frac{31}{36}\right)+\left(\frac{8}{9} \times \frac{31}{36}\right)^2+\cdots \cdot\right)\end{aligned} $
- the game then $B$ finishes it than the
- $
 \frac{1}{9}\left(\frac{1}{1-\frac{8}{9} \times \frac{31}{36}}\right)
$
- $=\frac{1}{9}\left(\frac{1}{1-\frac{248}{324}}\right)=\frac{1}{9}\left(\frac{1}{\frac{324-243}{324}}\right)$

- $
 =\frac{1}{9}\left(\frac{324}{76}\right)
 $
- $
 =\frac{36}{76}=\frac{9}{19}
 $
- [ If A starts the game then finishes it has the probability = $\frac{10}{19}$ ]

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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 L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- We know that

- $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$

- $\therefore 0.75 \leq P(A) + P(B) + P(C) + P(A \cap B \cap C) - P(A \cap B) - P(A \cap C) - P(B \cap C) \leq 1$

- or, $0.75 \leq 0.3 + 0.4 + 0.8 + 0.09 - 0.19 - 0.2 - P(B \cap C) \leq 1$

- or, $0.75 \leq 1.2 - P(B \cap C) \leq 1$

- or, $ -1 \leq P(B \cap C) - 1.2 \leq -0.75$

- $\therefore 0.2 \leq P(B \cap C) \leq 0.45$

- $\therefore$ Min. \& Max. values for $P(B \cap C)$ are 0.2 \& 0.45.

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

### <Success style="font-size:25px"> Probability L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- It is clear that $S \subseteq [0, 1] \times [0, 1]$

- Now  given that

- $ 0.5 \leq P(A \cup B) \leq 1 $(obvious)

- $ 0.125 \leq P(A \cap B) \leq 0.375 $

- $\therefore$ $ 0.625 \leq P(A \cup B + P(A \cap B) \leq 1.375$

- Now $P(A \cup B)+P(A \cap B)=P(A)+P(B)$

- $\therefore 0.625 \leq P(A) + P(B) \leq 1.375$

- or, $\frac{5}{8} \leq x + y \leq \frac{11}{8}$

- $\therefore ~ ~ $ S in PXYSRQ

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![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/000358b8-7511-4b5c-1695-a6d17dde1a00/public)

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

### <Success style="font-size:25px"> Probability L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Suppos $|A \cap B|=x \quad \therefore \quad P(A \cap B)=\frac{x}{6}$ now

- $ P(A)=\frac{|A|}{6} \quad$ & $P(B)=\frac{|B|}{6} $

- $ \frac{x}{6}=\frac{|A|}{6} \times \frac{|B|}{6} \Rightarrow 6 x=|A| \cdot|B| $

- $\therefore A$ and $B$ should be such that $|A| \times|B|$ is a multhple of 6

- If $|A|=6$ then $|B|$ can be
- $
 1,2,3,4,5
 $

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

### <Success style="font-size:25px"> Probability L-4 </Success>
### <primary style="font-size:24px"> Example </primary>
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- Consider rolling a fair die

- $\therefore P(1) = P(2) = ... = P(6)$

- $\therefore P(2) = \frac{1}{6}$

- Let A be the event that '2' has occured B be the event that an even number has occured

- $\therefore$ If B is known to have occured, then $P(A) = P(2)$ given that one of (2, 4 or 6) has occured

- $\therefore$ P(2 | even number has occured) $= \frac{1}{3}$

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

### <Success style="font-size:25px"> Probability L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \therefore P\left(G_3 \mid R_1 R_2 \cup G_1 G_2\right) $

- $ \quad \frac{12}{20} \times \frac{11}{19} \times \frac{8}{18}+\frac{8}{20} \times \frac{7}{19} \times \frac{6}{18} $

- $=\frac{12 \times 83+8 \times 42}{20 \times 19 \times 18}=\frac{27(132+42)}{70 \times 19 \times 13}=\frac{174}{19 \times 5 \times 5}=\frac{58}{19 \times 15} $

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![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/79b82db5-42d5-4547-f953-503e412d6300/public)

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