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?

Let $E_1$ be the event that the sum of faces $= 9$

This can happen with the following cases:

$(3, 6), (4, 5), (5, 4) \ \ \& \ \ (6, 3)$

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

Let $E_2$ be the event that the sum is $6$.

$\therefore$ Possibilities are $(1, 5), (2, 4), (3, 3), (4, 2) (5, 1)$
$\therefore P(E_2) = \frac{5}{36}$

We know that A starts the game

$\therefore$ A finishes the game is

• A throws 9 first time : $\frac{1}{9}$

• A throws other than 9 then B throws other than 6 then A thrown 9

  • Probability $= \frac{8}{9} \times \frac{31}{36} \times \frac{1}{9}$

$\frac{8}{9} \times \frac{31}{26} \times \frac {8}{9} \times \frac{31}{36} \times \frac{1}{9}$

$\frac{1}{9} \times (\frac{8 \times 31}{9 \times 36})^2$

$\frac{1}{9} + \frac{1}{9} \cdot (\frac{8}{9} \times \frac{31}{36}) + \frac{1}{9} (\frac{8}{9} \times \frac{31}{36})^2 +….$

$\frac{1}{9} (1+(\frac{8}{9} \times \frac{31}{36})+(\frac{8}{9} \times \frac{31}{36})^2+….)$

$=\frac{1}{9} (\frac{1}{1- \frac{8}{9} \times \frac{31}{36}})$

$= \frac{1}{9} (\frac{1}{1- \frac{248}{324}}) = \frac{1}{9} (\frac{1}{\frac{324 - 248}{324}})$

$= \frac{1}{9} (\frac{324}{76})=\frac{36}{76} = \frac{9}{19}$
$\therefore$ If A starts the game then B finishes it.
Then 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 P X Y S R Q

Note that length of PQ = $\sqrt{(\frac{5}{8})^2 + (\frac{5}{8})^2} = \sqrt{2 \times (\frac{5}{8})^2} = \sqrt{2}(\frac{5}{8})$

& length of XY = $\sqrt{(1-\frac{3}{8})^2 + (1-\frac{3}{8})^2} = \sqrt{2 \times (\frac{5}{8})^2} = \sqrt{2}(\frac{5}{8})$

Parimeter of S is the sum of 6 line segment:

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

$= 1.41 \times \frac{10}{8} + \frac{12}{8} = \frac{26.1}{8} \approx 3.26$ perimeter length

Area of S

$\frac{1}{2} \times (\frac{11}{8})^2 - \frac{1}{2} (\frac{5}{8})^2 - \frac{1}{2}(\frac{3}{8})^2 - \frac{1}{2}(\frac{3}{6})^2$

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

$\therefore$ The answer in perimeter = 3.26, 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 that, A and B are independent if $P(A \cap B) = P(A).P(B)$

B = {1, 2} A = {1, 2, 3}

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

$P(A \cup B)$ = P({1,2}) $= \frac{1}{3}$

$\therefore P(A \cup B) \neq P(A) \cdot P(B)$

$\therefore$ In this case A and B are not independent

Consider $\therefore$ B = {1, 2} A = {1, 3, 4}

$\therefore P(B) = \frac{1}{3}$ $P(A) = \frac{1}{2}$ $P(A \cap B)$ = P({1}) $= \frac{1}{6}$

$\therefore P(A \cap B) = P(B) \times P(A)$

Suppose $|A \cap B| = x$
$\therefore P(A \cap B) = \frac{x}{6}$

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

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

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

$\therefore |A| = 6$ than |B| can be 1, 2, 3, 4, 5

$\therefore$ possible cases are ${ }^6 c_1 + { }^6 c_2 + { }^6 c_3 + { }^6 c_4 + { }^6 c_5$

$=6 + \frac{6!}{2!4!} + \frac{6!}{3!3!} + \frac{6!}{4!2!} + \frac{6!}{5!1!}$

$=6 + 15 +20 +15 +6$

$= 62$ ways

$\therefore$ If |A| = 4, then

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

${ }^6 c_4 \times { }^4 c_2 \times { }^2 c_1 = 15 \times 6 \times 2 = 180$

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

$\therefore$ No. of ways is

${ }^6 c_3 \times { }^3 c_1 \times { }^3 c_1 = 20 \times 3 \times 3 =180$

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

When A & B are independent $= 62 + 180 + 180 = 422$

Suppose A and B are two events which are not independent. In that case the occurence or non - occurence of B may have an effect on the occurence or non - occurence of A consequents the P(A) given that B has occured will be different P(A) without the knowledge of occurence of B.

$\therefore P(1) = P(2) = … = P(6)$

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

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

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 take out 3 balls in succession without replacement.

We want to the conditional probability that the $2^{nd}$ ball is green given that the $1^{st}$ ball is red.

Let $R_1$ be the event that the first ball is red and $G_2$ be the event that the second ball is green.’

$\therefore$ We are looking $P(G_2 | R_1)$

$12 R + 8 G$

$\therefore P(G_1 | R_1) = \frac{8}{2}$

Now we know that $P(A|B) = \frac{8}{9}$

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

i.e., $P(R_1 \cap G_2) = P(G_2|R_1) \cdot (R_1) = \frac{8}{19} \times \frac{12}{20} = \frac{8}{19} \times \frac{3}{5}$

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

Solution

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

$\therefore$ Either both of them are red or both of them are green.

$\therefore P(G_3 | R_1 R_2) + P(G_3 | G_1 G_2)$

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

$=\frac{12 \times 88 + 8 \times 42}{20 \times 19 \times 18}= \frac{8(132 + 42)}{20 \times 19 \times 18}\\ = \frac{174}{19 \times 9 \times 5} = \frac{58}{19 \times 15}$