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.

For a specific game
let A_{i} (B_{i}) 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, A_{1} and B_{0} occur; A_{2} and B_{0}
occur, or A_{2} and B_{1} occur. If A_{W} be the event
that A wins the specific game, then .

P(A_{W}) =
P(A_{1} Ç B_{0}) +
P(A_{2} Ç B_{0}) +
P(A_{2} Ç B_{1})

= P(A_{1})×P(B_{0}) + P(A_{2})×P(B_{0}) + P(A_{2})×P(B_{1})

=

Now game results in
a tie if A_{0} and B_{0} occur or A_{1} and B_{1}
occur or A_{2} and B_{2} occur. If T be event representing a
tie;.

&⇒ P(T) = P(A_{0})×P(B_{0}) + P(A_{1}) ×P(B_{1}) + P(A_{2})×P(B_{2})

=

Now required probability,

= P(A_{W})
+ P(T)×P(A_{W}) +
(P(T))^{2}×P(A_{W}) + L

=

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.

Total number of ways of drawing the balls

= ^{2n}C_{2
}× ^{2n-2}C_{2
}× ^{2n}^{-}^{4}C_{2}L^{4}C_{2 }× ^{2}C_{2}

=

=

For favourable ways we must draw balls in pairs.

Number of choices
for first pair = n×n = n^{2}

Similarly number of choices for second pair

= (n -1) × (n -1)
= (n - 1)^{2}

similarly for the remaining pairs.

&⇒ Total favourable ways = n^{2}
× (n -1)^{2}×(n -2)^{2}×
L 2^{2}×1^{2 }= (n!)^{2}

&⇒ Probability of the required event is
=

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.

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 =

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.

Let A represents the event that sample taken does not posses any narcotic tablet. .

&⇒ P(A) =

&⇒ Probability that traveler would be
arrested =

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 r^{th} draw.

Procedure of drawing
the balls has to end at the r^{th} 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 r^{th} draw =

&⇒
Probability of required event =

=