Question

You enter a chess tournament where your probability of winning a game is $0.3$ against half the players (call them type $1$), $0.4$ against a quarter of the players (call them type $2$), and $0.5$ against the remaining quarter of the players (call them type $3$). You play a game against a randomly chosen opponent. What is the probability of winning?

• A. $0.375$
• B. $0.986$
• C. $0.236$
• D. $0.135$

Answer 1

The correct option is A $0.375$

Let $A_i$ be the event of playing an opponent of type $i$.

$A_1$ = playing the half of the opponent pool where the probability of winning $P\left(winning|A_1\right)=0.3$

$A_2$ = playing the half of the opponent pool type 2, where the probability of winning $P\left(winning|A_2\right)=0.4$

$A_3$ = playing the half of the opponent pool type 3, where the probability of winning $P\left(winning|A_3\right)=0.5$

By total probability theorem, we get,

$P\left(winning\right)=P\left(A_1\right)P\left(winning|A_1\right)+P\left(A_2\right)P\left(winning|A_2\right)+P\left(A_3\right)P\left(winning|A_3\right)$

$P\left(winning\right)=50\%\times 0.3+25\%\times 0.4+25\%\times 0.5$

$P\left(winning\right)=\mathrm{0.375.}$