• $P(A)$, where $A$ is an event corresponding to a random experment $E$ whose sample space is $\Omega$, is $\frac{\#A}{\#\Omega}.$
• $P(A \cup B) = P(A) + P(B)$ if $A$ and $B$ are disjoint.
• But if $A$ & $B$ are not disjoint then
  $P(A \cup B) = P(A) + P(B) - P(A \cap B).$
• If $A$ & $B$ are independent events then
  $P(A \cap B) = P(A) \times P(B).$

If $x, y, z \geqslant 0$ and $x + y + z = 10$ then number of possible solution is

$^{10+3-1}C_{3-1} = ^{12}C_2=66$

What is the probability that in the above scenario $X$ is an odd number?

$X$ can be odd $\Rightarrow X$ can take values $\{1, 3, 5, 7, 9\}$

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

$X = 3 \Rightarrow \ Y + Z = 7 : ………………………… = 8$

$X = 5 \Rightarrow Y + Z = 5 : …………………………= 6$

$X = 7 \Rightarrow Y + Z = 3 : …………………………= 4$

$X = 9 \rightarrow Y + Z = 1 : …………………………= 2$

Now, $10+8+6+4+2= 30$

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

Suppose a man is standing on the x-axis at the origin. At each step he goes either to the right or to the left each with probability $=\frac{1}{2}.$

What is the probability that after 6 steps he will be at a distance 2 from the origin ?

Let $X$ be the number of steps he takes to the right direction & $Y$ be the number of steps he takes to the left direction.

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

$\therefore$ Desired Probability $= 2(0.5)^6$

$\quad \quad \quad=(0.5)^5$

Suppose A, B, C are three events corresponding to a random experiment E, such that

(i) $P(A \cup B \cup C) = 1$

(ii) $A, B, \ \ \& \ \ C$ are equally likely.

(iii) $A \cap B, B \cap C \ \ \& \ \ A \cap C$ are also equally likely.

(iv) $P(A \cap B) = \frac{P(A)}{2} \ \ \& \ \ P(A \cap B \cap C) = \frac{P(A \cap B)}{2}$

Find the probability of $A \cap B \cap C^c.$

Let $P(A) = x$ $\ \ \therefore P(B) = P(C) = x$

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

and $P(A \cap B \cap C)=\frac{x}{4}$
$\therefore P(A \cup B \cup C) = P(A)+P(B)+P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$

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

$\Rightarrow 4 = 12x + x - 6x = 7x$

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

$=\frac{x}{2}-\frac{x}{4} =\frac{2}{7}-\frac{1}{7}$

$=\frac{1}{7}$

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 combination 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 $B$ be the event that the face on both the dice is even

$\therefore$ number of possibilities :

$(2,2) (2,4) (2,6)\\(4,2) (4,4) (4,6)\\(6,2) (6,4) (6,6)\biggl\} =9$

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

$\therefore P(A \cup B) =P(A)+P(B)-\underbrace{P(A \cap B)}_{=\frac{3}{36}}$

$=\frac{5}{36}+\frac{9}{36}-\frac{3}{36}=\frac{11}{36}$

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?

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

Let us computs the probabilites that A will win the tournament for both the option.

Best of 3 scenario

Here the player who wins two matches first is the winner.

$\therefore$ A can win if

(1) he wins both first and second match.

or

(2) he wins one of first and second match and then wins the third match.

$\therefore \ P(A \ \text{wins the tournament}) \ \text{if} \ \ WW, WLW, LWW$

$WW - (0.4)×(0.4)$

$WLW - (0.4)(0.6)(0.4)$

$LWW - (0.6)(0.4)(0.4)$

$P(A \ \text{wins the tournament}) \ =(0.4)^2+(0.4)^2 × 0.6+(0.4)^2 × 0.6$

$=(0.4)^2(1+0.6+0.6)$

$= (0.16) × (2.2) = 0.352$

Best of 5 scenario

Final can be over with A winning in the following ways:

(1) A wins the first 3 matches. $(0.4)^3$

(2) A wins two of the first 3 matches & then wins the $4^{th}$ match.

Its probability is.

$3.(0.4)^3 (0.6)$
$\{\because WWL, WLW, LWW \rightarrow (0.4)^2.(0.6).(0.4)\}$

(3) A wins two of the first 4 matches and then wins the $5^{th}$ one.

A wins 2 matches. This can be done in ${}^{4}C_2$ ways

$W W L L$

$W L W L$

$W L L W$

$L W W L$

$L W L W$

$L L W W$

Then wins $5^{\text{th}}$ match $= W$

$\therefore$ The probability is :

$6\times(0.4)^2 \times(0.6)^2 \cdot(0.4)$

$(0.4)^3 \times6^2 \cdot(0.4)$

$\therefore$ P(A wins the tournament)

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

$=(0.4)^3(1+1.8+2.16)$

$\approx 0.317 < 0.352$

A should choose best of 3 option.

Suppose 5 candidates A, B, C, D & E are waiting to appear in an interview. It is known that the interview board will call them randomly in any order

a) Find the probability that A is called before B.

b) Find the probability that A comes before B & B is called before C.

C) Find the probabality that B is called just after A.

a) A is called before B.

• A is called at no 1. $\ \$ A _ _ _ _

  $\therefore$ Number of possibility $= 4!$

• A is called at no 2. $\ \$ _ A _ _ _

  $\therefore$ Number of possibility $= 3×3!$

• A is called at no 3. $\ \$ _ _ A _ _

  $\therefore 3×2×2 = 12$

• A is called at no 4. $\ \$ _ _ _ A _

$\therefore 3! = 6$

$\therefore$ Number of possibility $\ = 24 + 18 + 12 + 6 = 60$

Now total number of parmutations of $5$ number is $5! = 120$
$\therefore \ P(A \ \text{is called before} \ B) \ = \frac{60}{120}=\frac{1}{2}$

(b) A before B & B before C.

• A is in position 1.

  • B is in $2^{nd}$ position. Then number of posibility $= 6$
  • B is in $3^{rd}$ position. Then number of posibility $2 × 2 = 4$
  • B is in $4^{th}$ position. Then number of posibility $=2$
    Total number of posibility $=6+4+2=12$

• A is in position 2.

  • B is in $3^{rd}$ position. Then number of posibility $2 × 2 = 4$
  • B is in $4^{th}$ position. Then number of posibility $=6$

A is in $3^{rd}$ position.

$\therefore$ D & E are in $1^{st}$ & $2^{nd}$ position that can be done in 2 ways.

$\therefore$ Total possibilities $= 12 + 6 + 2 = 20$

$\therefore \ P(A \ \text{comes before} \ B, B \ \text{before} \ C)$

$=\frac{20}{120}=\frac{1}{6}$

C) A comes just before B.

$\therefore$ AB as one unit &

$\therefore$ We have now AB, C, D, E and they can be arranged in $4! = 24$ ways.

$\therefore$ Probability = $\frac{24}{120} =\frac{1}{5}$

Suppose $5$ persons A, B, C, D & E are sitting in a circular table you have hats of $3$ colors: W, R & G.

In how many ways you can give one hat to each person, $\ni$ no two consecutive person have hats of same colour?

But A can get W, G & R.

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

Suppose two teams MB and EB are facing each other in two matches. For each match the winner gets 3 points & loser get 1 point for a drawn match both the teams get 1 point each.

If MB has .5 probability of winning,

.1 probability if drawning,

& .4 probability of losing

What is the probability that MB ends with more points than EB?

$\therefore$ P(MB > EB)

$= .5 × .5 + .5 × .1 + .5 × .1 \\ = .25 + .05 + .05 \\ = .25 + .1 = .35$

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

$.2 + .01 + .2 = 0.41$

$\therefore$ P(MB < EB)

$= 0.24$