2_Topic_Q27

Question

A and B a set of games. The chance of winning the next game is 2:1 in favour of the winner from the previous game. If A wins the first game, what are the chances that A wins at least three of the next four games?

(A) 4/9

(B) 2/9

(C) 1/3

(D) none

Comments

D ?

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Sorry, sir @Shobhit Joshi but its A in the answer key....

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@BOB i did a calculation mistake, my bad and no need to call me sir.

Answer 1

Case 1: When A wins the next 3 games only :

B(lose)*A(win)*A(lose)*A(win)+

A(win)*B(lose)*A(lose)*A(win)+

A(win)*A(win)*B(lose)*A(lose)+

A(win)*A(win)*A(win)*B(lose)

A(win) = A win this round and was winnner in the previous round

A(lose) = A wins this round and was the loser

= $\frac{1}{3}*\frac{1}{3}*\frac{2}{3}*\frac{2}{3}+\frac{2}{3}*\frac{1}{3}*\frac{1}{3}*\frac{2}{3}+\frac{2}{3}*\frac{2}{3}*\frac{1}{3}*\frac{1}{3}+\frac{2}{3}*\frac{2}{3}*\frac{2}{3}*\frac{1}{3}=\frac{20}{81}$

Case 2: When A wins all the four games: A(win)*A(win)*A(win)*A(win)=$\frac{2}{3}*\frac{2}{3}*\frac{2}{3}*\frac{2}{3}=\frac{16}{81}$

required probability = $\frac{20}{81}+\frac{16}{81}=\frac{4}{9}$

Correc me if I'm wrong

Comments

Thanks, @Shobhit Joshi... :)