JEE MAIN - Mathematics (2018 - 15th April Evening Slot - No. 10)
A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If X starts the game, and the probability of winning the game by both the players is equal, then the value of 'p' is :
$${1 \over 5}$$
$${1 \over 3}$$
$${2 \over 5}$$
$${1 \over 4}$$
Explanation
P(X getting head) = p
$$ \therefore $$ P(X getting tail) = 1 - p
P(Y getting head) = P(Y getting tail) = $${1 \over 2}$$
P(X wins) = p + (1 - p)$${1 \over 2}$$p + (1 - p)$${1 \over 2}$$(1 - p)$${1 \over 2}$$p + ...
= $${p \over {1 - \left( {{{1 - p} \over 2}} \right)}}$$
= $${{2p} \over {1 + p}}$$
P(Y win) = (1 - p)$${1 \over 2}$$ + (1 - p)$${1 \over 2}$$(1 - p)$${1 \over 2}$$ + ...
= $$\left( {{{1 - p} \over 2}} \right).{p \over {1 - \left( {{{1 - p} \over 2}} \right)}} = {{1 - p} \over {1 + p}}$$
According to question,
P(X wins) = P(Y wins)
$$ \therefore $$ $${{2p} \over {1 + p}}$$ = $${{1 - p} \over {1 + p}}$$
$$ \Rightarrow $$ 3p = 1
$$ \Rightarrow $$ p = $${1 \over 3}$$
$$ \therefore $$ P(X getting tail) = 1 - p
P(Y getting head) = P(Y getting tail) = $${1 \over 2}$$
P(X wins) = p + (1 - p)$${1 \over 2}$$p + (1 - p)$${1 \over 2}$$(1 - p)$${1 \over 2}$$p + ...
= $${p \over {1 - \left( {{{1 - p} \over 2}} \right)}}$$
= $${{2p} \over {1 + p}}$$
P(Y win) = (1 - p)$${1 \over 2}$$ + (1 - p)$${1 \over 2}$$(1 - p)$${1 \over 2}$$ + ...
= $$\left( {{{1 - p} \over 2}} \right).{p \over {1 - \left( {{{1 - p} \over 2}} \right)}} = {{1 - p} \over {1 + p}}$$
According to question,
P(X wins) = P(Y wins)
$$ \therefore $$ $${{2p} \over {1 + p}}$$ = $${{1 - p} \over {1 + p}}$$
$$ \Rightarrow $$ 3p = 1
$$ \Rightarrow $$ p = $${1 \over 3}$$
Comments (0)
