JEE Advanced 2022 Paper 2 · Mathematics · Probability · medium

Players A and B throw a fair die alternately, A going first, and the first player to throw a six wins. Given that the game is decided within the…

Numerical answer

Players

A

and

B

throw a fair die alternately,

A

going first, and the first player to throw a six wins. Given that the game is decided within the first four throws, let

p

be the probability that

B

is the winner. The value of

11p

Show answer and solution

Answer: 5

Let

q=5/6

. The game ends on throw

k

with probability

\frac16 q^{k-1}

. It is decided within four throws with probability

\frac16(1+q+q^2+q^3)

. Player

B

wins only on even throws

2

and

4

, with total probability

\frac16(q+q^3)

. The conditional probability is

p=\dfrac{q+q^{3}}{1+q+q^{2}+q^{3}}=\dfrac{q(1+q^{2})}{(1+q)(1+q^{2})}=\dfrac{q}{1+q}=\dfrac{5/6}{11/6}=\dfrac5{11}

. Therefore

11p=5

. The unconditional answer for an unbounded game is also

5/11

, since the same cancellation of

(1+q^2)

occurs there.

Original question written in the JEE Advanced 2022 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake