Example.1
A tosses 2 fair coins and B tosses 3 fair
coins. The game is won by the person who throws greater number of heads. In
case of a tie, the game is continued under identical rules until someone wins
the game. Find the probability of A winning the game.
Solution
For a specific game
let Ai (Bi) denote the number of heads obtained by A(B)
is i when he tosses two (three) fair coins. A will win a particular game in
following ways, A1 and B0 occur; A2 and B0
occur, or A2 and B1 occur. If AW be the event
that A wins the specific game, then .
P(AW) =
P(A1 Ç B0) +
P(A2 Ç B0) +
P(A2 Ç B1)
= P(A1)×P(B0) + P(A2)×P(B0) + P(A2)×P(B1)
= 
Now game results in
a tie if A0 and B0 occur or A1 and B1
occur or A2 and B2 occur. If T be event representing a
tie;.
&⇒ P(T) = P(A0)×P(B0) + P(A1) ×P(B1) + P(A2)×P(B2)
= 
Now required
probability,
= P(AW)
+ P(T)×P(AW) +
(P(T))2×P(AW) + L
= 
Example.2
A bag contains n white and n red balls.
Pairs of balls are drawn without replacement until the bag is empty. Find the
probability of each pair consisting of balls of different colours.
Solution
Total number of ways
of drawing the balls
= 2nC2
× 2n-2C2
× 2n-4C2L4C2 × 2C2
= 
= 
For favourable ways
we must draw balls in pairs.
Number of choices
for first pair = n×n = n2
Similarly number of
choices for second pair
= (n -1) × (n -1)
= (n - 1)2
similarly for the
remaining pairs.
&⇒ Total favourable ways = n2
× (n -1)2×(n -2)2×
L 22×12 = (n!)2
&⇒ Probability of the required event is
=
Example.3
A draws a card from
a pack of n cards marked 1, 2, .... ,n. The card is replaced in the pack and B
draws a card. Find the probability that A draws (i) the same card as B, (ii) a
higher card than B.
Solution
A and B can draw the cards in n ´ n ways. .
(i) If A draws the same card as B then
number of favourable cases in this is n.
Therefore, required probability = 
(ii) If A draws card higher than B then
number of favourable cases is
(n - 1) + (n - 2) +L + 3 + 2 + 1.
(As when B draws card number 1 then A can
draw any card from 2 to n and so on).
Therefore, required probability = 
Example.4
To avoid detection at customs, a traveler
has placed six narcotic tablets in a bottle containing nine vitamin pills that
are similar in appearance. If the customs official selects three of the tablets
at random for analysis, what is the probability that traveler will be arrested
for illegal possession of narcotics.
Solution
Let A represents the
event that sample taken does not posses any narcotic tablet. .
&⇒ P(A) = 
&⇒ Probability that traveler would be
arrested = 
Example.5
A bag contains ‘W’ white and 3 black
balls. Balls are drawn one by one without replacement till all the black balls
are drawn. What is the probability that this procedure for drawing the balls
will come to an end at the rth draw.
Solution
Procedure of drawing
the balls has to end at the rth draw,
&⇒
exactly 2 black balls must have been drawn up to (r -1)th draw.
Now probability of drawing exactly 2 black
balls up to (r -1) draws
=
= 
At the end of (r -1)th draw, we would be
left 1 black and (W -r+3) white balls. .
&⇒
Probability of drawing the black ball at the rth draw = 
&⇒
Probability of required event = 
= 
