A plays three games against B. The games are independent of one another, no game ends in a draw, and A wins each game with probability \(\displaystyle\frac{3}{5}\). If it is known that A won exactly two of the three games, then the probability that A won the first game is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{2}{3}\)
Write W for a win of A and L for a loss. A wins exactly two games in three ways: WWL, WLW, LWW. The games are independent, so each way has the same probability:
\( \displaystyle \frac{3}{5}\times\frac{3}{5}\times\frac{2}{5} = \frac{18}{125} \)
So
\( \displaystyle P(\text{exactly two wins}) = 3\times\frac{18}{125} = \frac{54}{125} \)
A won the first game in two of these ways, WWL and WLW:
\( \displaystyle P(\text{first won and exactly two wins}) = 2\times\frac{18}{125} = \frac{36}{125} \)
By the definition of conditional probability:
\( \displaystyle P = \frac{36}{125}\div\frac{54}{125} = \frac{2}{3} \)
The value \(\frac{3}{5}\) cancels out: given exactly two wins, the three ways are equally likely, and A won the first game in 2 of them.
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{5}\) — Incorrect. This is the chance of winning the first game when nothing is known about the other games; the condition changes it.
Option (3) \(\displaystyle\frac{9}{25}\) — Incorrect. This is \(\frac{3}{5}\times\frac{3}{5}\), the probability of winning the first two games.
Option (4) \(\displaystyle\frac{1}{3}\) — Incorrect. This is the probability that A lost the first game, given exactly two wins (only LWW).