mathematics-class-12-unit-13-chapter-13-probability-l-3-6

### <Success style="font-size:25px"> Probability L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- X can be odd $\Rightarrow$ X can take values {1, 3, 5, 7, 9}

- $x=1 \Rightarrow Y+Z = 9$ : (0, 9), (1, 8), (2, 7) ... (9, 0) = 10

- $x=3 \Rightarrow Y+Z = 7$ : (0, 9), (1, 8), (2, 7) ... (9, 0) = 8

- $x=5 \Rightarrow Y+Z = 5$ : (0, 9), (1, 8), (2, 7) ... (9, 0) = 6

- $x=7 \Rightarrow Y+Z = 3$ : (0, 9), (1, 8), (2, 7) ... (9, 0) = 4

- $x=9 \Rightarrow Y+Z = 1$ : (0, 9), (1, 8), (2, 7) ... (9, 0) = 2

- = 30

- $\therefore$ P(X is an odd) $= \frac{30}{66} = \frac {5}{11}$

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

### <Success style="font-size:25px"> Probability L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Let X be the number of steps he takes to the right direction and Y be the number of steps he takes to the left direction.

- $X+Y=6$ \& $|X-Y|=2$

- We get,
- i) $X+Y=6$

- $X-Y=2$,  $\Rightarrow 2X=8$

- $\Rightarrow X=4$,  $\Rightarrow Y=2$,  $=(0.5)^6$

- ii) $X+Y=6$, $Y-X=2$

- $\Rightarrow 2Y=8$, $\Rightarrow Y=4$ ,$\Rightarrow X=2$,  $=(0.5)^6$

- $\therefore$ Desired Probabililty

- $=2(0.5)^6$, $=(0.5)^5$
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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-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $P(A) = x$

- $\therefore P(A \cap C)=\frac{P(A)}{2}=\frac{x}{2}$
\& $P(A \cap B \cap C) = \frac{x}{4}$
- $P(B) $

- $= P(C) = x$
- $=P(B \cap C)=P(A \cap C)$

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

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- $\Rightarrow 1 = 3x+\frac{x}{4}-\frac{3x}{2}$

- $\Rightarrow 4 = 12x + x - 6x$
- $=7x$$\qquad$
- $\therefore x =\frac{4}{7}$

- $\therefore P(A \cap B \cap C^{c})=P(A \cap B) - P(A \cup B \cup C)$
- $=\frac{x}{2}-\frac{x}{4}$
- $=\frac{2}{7}-\frac{1}{7}$
- $=\frac{1}{7}$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Probability L-3 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- Suppose a pair of dice is rolled. The dice is fair i.e. P(1) = P(2) = ... P(6) = $\frac{1}{6}$
- What is the probability that the sum is 8 or you get even numbers on both the dice?

- Solution:

- Let A be the event that the sum of the two faces is 8

- $\therefore$ Possible combinations are: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) = 5$

- $\therefore P(A) = \frac{5}{36}$  $\because$ There are 36 pair of points in $\Omega$.
- Let B is the event that the face on both the dice is even
\# of possibilities :

- (2, 2)  (2, 4)  (2, 6) (4, 2)  (4, 4)  (4, 6) (6, 2)  (6, 4)  (4, 6)$ \quad =9$

- $\therefore P(B) = \frac{9}{36}$

- $\therefore P(A \cup B) = P(A) + P(B) - \frac{P(A \cap B)}{3/36}$
- $= \frac{5}{36} + \frac{9}{36} - \frac{3}{36} = \frac{11}{36}$

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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-3 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
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- Suppose A and B are playing the final of a tournament against each other. The final may be best of 3 matches or best of 5 matches. The probability that A wins a match against B is 0.4

- But A has the advantage that he can choose whether it will best of 3 or best of 5

- Which one will A choose?

- Solution:

- Obviously A will choose format where he has more chance of winning.

- Let us complete the probabilities that A will win the tournament for both the option.
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Best of 3 scenario
- Here the player who wins two matches first in the winner.

- $\therefore$ A can win if 
- or,
- 1) He wins both first and second match,$\quad$
- 2) He wins one of first and second match and then wins the third match
- P (A wins the tournament) if
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|     W    |     W    |
| -------- | -------- |
|    0.4   |    0.4   |,

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|     W    |     L    |     W    |
| -------- | -------- | -------- |
|    0.4   |    0.6   |    0.4   |, 
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|     L    |     W    |     W    |
| -------- | -------- | -------- |
|    0.6   |    0.4   |    0.4   |, 
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- $(0.4)^2 + (0.4)^2 \times 0.6 + (0.4)^2 \times 0.6$
- $= (0.4)^2 (1 + 0.6 + 0.6)$
- $= (0.16) \times (2.2) = 0.352$

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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 L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $W : (0.4)^2 \times (0.6) \times (0.4)$
- A wins 2

- This can be done in ${ }^4 c_2$ ways

- $\therefore$ The probabilities is:

- $6 \times (0.4)^2 \times (0.6)^2 \times (0.4)$
- $= (0.4)^3 \times 6 \times (0.6)^2$

- $\therefore$ P(A wins the tournament)

- = $(0.4)^3 + 3 (0.4)^3 \times 0.6 + 6 \times (0.6)^2 \times (0.4)^3$

- $= (0.4)^3 (1 + 1.8 +2.16)$
- $\approx 0.317 < 0.352$

- A should choose best of 3 option.

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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 L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- a) A is called before B
- A is called at no. 1
- $\quad \therefore$ No. of possibilities = 4
- A is called at no. 2
- $\quad\therefore$ No. of possibilities = $3 \times 3!$
- A is called at no. 3
- $\quad\therefore 3 \times 2 \times 2 = 12$
- A is called at no. 4
- $\quad\therefore 3! = 6$ $\therefore$ \# of possibilities $= 24 + 18 + 12 + 6=60$

- Now total no. of possibilities of 5 no. is $5!=120$

- $\therefore$ P(A is called before B) $=\frac{60}{120}=\frac{1}{2}$

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

### <Success style="font-size:25px"> Probability L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- (b) A before B & B before C

- B is in $2^{nd}$ position

- $\therefore$ possibilities = 6
- B is in $3^{rd}$ position
- $2 \times 2 = 4$
- B is in $4^th$ position
- There are 2 possibilities
- A is in position 2 $\Rightarrow$ 12 possibilities

- $\therefore$ B is in $3^{rd}$ : $2 \times 2 = 4$
- $\therefore$ B is in $4^{th}$ : $2 = 2$
- $=6$

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- $\therefore$ A is in $3^{rd}$ POsition: D \& E are in $1^{st}$ \7 $2^{nd}$ position that can be done in 2 ways.

- $\therefore$ Total possibities $= 12 + 6 + 2 = 20$

- $\therefore$ P(A comes before B, B before C)
- $=\frac{20}{120}=\frac{1}{6}$

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<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Probability L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- But A can get W, G & R.

- $\therefore$ Total possibilities $= 10 × 3 = 30$

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![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/665d548c-d629-4f24-8031-33cac809b800/public)

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-7 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-8 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\therefore P(MB>EB)$

- $=0\cdot 5 \times 0 \cdot 5 + 0 \cdot 5 \times 0 \cdot 1 + 0 \cdot 5 \times 0 \cdot 1$

- $= 0 \cdot 25 + 0 \cdot 05 + 0 \cdot 05$
- $= 0 \cdot 25 + 0 \cdot 1 = 0 \cdot 35$

- What is the probability that they end up with same point

- $ 0\cdot 2 + 0 \cdot 01 + 0 \cdot 2$
- $=0 \cdot 41$

- $\therefore P(MB<EB)$
- $=0.24$
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-8 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>