In a match between two players X and Y, the player who first wins two games wins the match, so at most three games are played. No game is drawn, and X wins each game with probability 0.6, independently of the other games. The probability that X wins the match is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.648
Write W for a game won by X and L for a game lost by X. X wins the match in exactly three ways: WW, WLW and LWW. X loses a game with probability 0.4, and the games are independent, so
\(P(WW) = 0.6\times 0.6 = 0.36\)
\(P(WLW) = 0.6\times 0.4\times 0.6 = 0.144\)
\(P(LWW) = 0.4\times 0.6\times 0.6 = 0.144\)
The three ways cannot happen together, so their probabilities add:
\(P(\text{X wins the match}) = 0.36+0.144+0.144 = 0.648\)
Why the other options are incorrect
Option (1) 0.36 — Incorrect. This counts only the win in the first two games, WW.
Option (2) 0.288 — Incorrect. This counts only the wins that need three games, WLW and LWW.
Option (4) 0.432 — Incorrect. This is \(3\times 0.6\times 0.6\times 0.4\), the chance of exactly two wins if all three games were played; it misses three straight wins, \(0.6^{3} = 0.216\), and \(0.432+0.216 = 0.648\).