A player participates in 3 matches against three teams \(T_1, T_2\) and \(T_3\). The probability of winning a match against teams \(T_1, T_2\) and \(T_3\) are 0.2, 0.3 and 0.9 respectively. If ‘wins’ can be regarded as independent events, then the probability that he
(A) wins all the 3 matches is 0.054
(B) wins no match is 0.054
(C) wins exactly two matches is 0.348
(D) wins exactly one match is 0.542
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (C) and (D) only
The probabilities of winning are 0.2, 0.3, 0.9, so the probabilities of losing are 0.8, 0.7, 0.1. For independent events, the probability of a given result in all three matches is the product of the separate probabilities.
(A) wins all the 3 matches is 0.054 — Correct. All three wins:
\(0.2 \times 0.3 \times 0.9 = 0.054\)
(B) wins no match is 0.054 — Incorrect. All three losses give 0.056:
\(0.8 \times 0.7 \times 0.1 = 0.056\)
(C) wins exactly two matches is 0.348 — Correct. The lost match can be against \(T_3\), \(T_2\) or \(T_1\):
\(0.2 \times 0.3 \times 0.1 = 0.006\)
\(0.2 \times 0.7 \times 0.9 = 0.126\)
\(0.8 \times 0.3 \times 0.9 = 0.216\)
\(0.006 + 0.126 + 0.216 = 0.348\)
(D) wins exactly one match is 0.542 — Correct. The match won can be against \(T_1\), \(T_2\) or \(T_3\):
\(0.2 \times 0.7 \times 0.1 = 0.014\)
\(0.8 \times 0.3 \times 0.1 = 0.024\)
\(0.8 \times 0.7 \times 0.9 = 0.504\)
\(0.014 + 0.024 + 0.504 = 0.542\)
As a check, the four cases add up to 1: \(0.054 + 0.056 + 0.348 + 0.542 = 1\).