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.

$(3,6)(4,5),(5,4) (6,3)$.

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

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

$\frac{8}{9} \times \frac{31}{36} \times \frac{1}{9}$

$\begin{aligned} & \frac{8}{9} \times \frac{31}{36} \times \frac{8}{9} \times \frac{31}{36} \times \frac{1}{9} \ = & \frac{1}{9} \times\left(\frac{8 \times 31}{9 \times 36}\right)^2 \end{aligned}$

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

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

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 = $\frac{10}{19}$ ]

Let A, B, C be three events

$P(A) = 0.3 P(B) = 0.4 P(C) = 0.8$

$P(A \cup B) = 0.19 \quad P(A \cup C) = 0.2 \quad P(A \cup B \cup C) = 0.09$

If $P(A \cap B \cap C) \geq 0.75$

Find the minimum and maximum values for $P(B \cup C)$.

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.

Let A and B be two events of a random experiment E. It is given that

i) $P(A \cup B) \geq 0.5$

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

$\therefore$ There are many different probabilities are possible for A & B satisfying the above.

Let S be the region on 2D - plane

$S=(x,y)$ where x is the P(A) & y is the P(B) satisfying the above conditions

Find the area & Perimeter of S.

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

Note that length of $P Q=\sqrt{\left(\frac{5}{8}\right)^2+\left(\frac{5}{8}\right)^2}=\sqrt{2 \times\left(\frac{5}{8}\right)^2}=\sqrt{2}\left(\frac{5}{8}\right)$

and length of $x y=\sqrt{\left(1-\frac{3}{8}\right)^2+\left(1-\frac{3}{8}\right)^2}=\sqrt{2 \times\left(\frac{5}{8}\right)^2}=\sqrt{2}\left(\frac{5}{8}\right)$

Perimeter of $S$ in the sum of the 6 line segments:

$2 \sqrt{2}\left(\frac{5}{8}\right)+\frac{3}{8} * 4 =\sqrt{2} \frac{10}{8}+\frac{12}{8}$

$=141 \times \frac{10}{8}+\frac{12}{8}=\frac{26.1}{8}$

$\approx 3 \cdot 26$

$\begin{aligned} & \frac{1}{2} \times\left(\frac{11}{8}\right)^2-\frac{1}{2}\left(\frac{5}{6}\right)^2- \frac{1}{2}\left(\frac{3}{6}\right)^2-\frac{1}{2}\left(\frac{3}{6}\right)^2 \end{aligned}$

$\begin{aligned} & =\frac{1}{2 \times 64}(121-25-9-9)=\frac{78}{128}=0.609 \end{aligned}$

The answer in perimeter $=3.25$

Area $=0.609$

Suppose the sample space of a random experiment in {1, 2, 3, 4, 5, 6}. Where each element is equally likely.

If A and B are two independent events then compute the number of ordered pairs

$(A, B) \ni 1 \leq |B| < |A|$

Note $A$ and $B$ are independent if

$P(A \cap B)=P(A) \cdot P(B) .$

$\begin{aligned}& B=\{1,2\} \quad A=\{1,2,3\}\end{aligned}$

$\therefore ~ \begin{aligned} & P(B)=\frac{1}{3} \quad P(A)=\frac{1}{2} \quad P(A \cap B)=P(\{1,2\}) \end{aligned}$

$\therefore ~ \begin{aligned}& \quad P(A \cap B) \neq P(A) \cdot P(B)=\frac{1}{3}\end{aligned}$

In thin case $A$ and $B$ are not independent

Consider

$~ \therefore \quad B=\{1,2\} \quad A-\{1,3,4\}$

$\therefore P(B)=\frac{1}{3} \quad P(A)=\frac{1}{2} \quad P(A \cap B)=P(\{1\})$

& $~ ~ \therefore ~ P(A \cap B)={P(B) \times P(A)}=\frac{1}{6}$

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

$1,2,3,4,5$

$\therefore$ possible cases are

$\quad 6_{c_1}+6_{c_2}+6_{c_3}+6_{c_4}+6_{c_5}$

$\begin{aligned} & =6+\frac{6 !}{2 ! 4 !}+\frac{6 !}{3 ! 3 !}+\frac{6 !}{4 ! 2 !}+\frac{6 !}{5 ! 1 !} \ & =6+15+20+15+6 \ & =62 \text { aveg. } \end{aligned}$

If $|A|=4$ then

$|B|$ has to be 3 and $|A \cap B|$ has to be 2 .

possible cases are

${ }^6 C_4 \times{ }^4 C_2 \times{ }^2 C_1=15 \times 6 \times 2=180$

If $|A|=3$ then $|B|$ has is be 2 and $|A \cap B|=1$ element.

$\therefore$ No. of ways is

$\quad 6_{c_3}+3_{c_1}+3_{c_1} = 20 ×3×3 =180$

$\therefore$ Total number of possibilities $1 \leq|B|\langle|A|$

when $A$ & $B$ are independent $=62+180+180$

$=422$

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) $= \frac{1}{3}$

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

So in the above case $A \cap B =${2}

B = {2, 4, 6}

$\therefore P(A|B) = \frac{P(2)}{P(2,4,6)} = \frac{1/6}{1/2} = \frac{1}{3}$

Consider a bag containing 12 Red balls and 8 green balls.

We the out 3 ball in uncommon without replacement.

We want to the the conditional probabilities that the $2^{\text {nd }}$ ball in green given that the $1^{\text {st }}$ ball is red

Solution:

Let $R$, be the event that the first ball in rad. & $G_2$ be the event that the second ball in green

: We are booking $P\left(G_2 \mid R_1\right)$

$12 \mathrm{R} / \mathrm{t} \mathrm{G}$

Now she know that $P(A \mid B)$ = $\frac{P(A \cap B)}{P(B)}$

$\therefore$ We can write as: $P(A \cap B)=P(A \mid B) \cdot P(B)$

ie $P\left(R_1 \cap G_2\right)=P\left(G_2 \mid R_1\right) P\left(R_1\right)=\frac{8}{19} \times \frac{12}{20}=\frac{8}{19} \times \frac{3}{5}$

What is the probability that the $3^{\text {rd }}$ ball drawn in green given that the first two balls drawn are of the same colour?

The given condition in ball 1 and ball 2 drawm are of the same colour.

Solution:

$\therefore$ The given condition is ball 1 & ball 2 drawn are of the same colour.

$\therefore$ Either both of then are red or both of then are green

$\therefore P\left(G_3 \mid R_1 R_2\right)+P\left(G_3 \mid G_1 G_2\right)$

$\therefore P\left(G_3 \mid R_1 R_2 \cup G_1 G_2\right)$

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

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