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:

(A), (C) and (D) only

The correct answer is Option (1) → (A), (C) and (D) only

(A) wins all the 3 matches is 0.054 (Correct)
(B) wins no match is 0.054 (Incorrect)
(C) wins exactly two matches is 0.348 (Correct)
(D) wins exactly one match is 0.542 (Correct)

Let $P_1 = 0.2$, $P_2 = 0.3$, $P_3 = 0.9$ be the probabilities of winning against teams $T_1$, $T_2$, and $T_3$ respectively.

Then, $Q_1 = 1 - 0.2 = 0.8$, $Q_2 = 1 - 0.3 = 0.7$, $Q_3 = 1 - 0.9 = 0.1$ are the probabilities of losing each match.

(A) Probability of winning all 3 matches:

$P(\text{WWW}) = P_1 \cdot P_2 \cdot P_3 = 0.2 \cdot 0.3 \cdot 0.9 = 0.054$

(B) Probability of winning no match:

$P(\text{LLL}) = Q_1 \cdot Q_2 \cdot Q_3 = 0.8 \cdot 0.7 \cdot 0.1 = 0.056$

(C) Probability of winning exactly 2 matches:

$P(\text{WWL}) = P_1 \cdot P_2 \cdot Q_3 = 0.2 \cdot 0.3 \cdot 0.1 = 0.006$

$P(\text{WLW}) = P_1 \cdot Q_2 \cdot P_3 = 0.2 \cdot 0.7 \cdot 0.9 = 0.126$

$P(\text{LWW}) = Q_1 \cdot P_2 \cdot P_3 = 0.8 \cdot 0.3 \cdot 0.9 = 0.216$

Total = $0.006 + 0.126 + 0.216 = 0.348$

(D) Probability of winning exactly 1 match:

$P(\text{WLL}) = P_1 \cdot Q_2 \cdot Q_3 = 0.2 \cdot 0.7 \cdot 0.1 = 0.014$

$P(\text{LWL}) = Q_1 \cdot P_2 \cdot Q_3 = 0.8 \cdot 0.3 \cdot 0.1 = 0.024$

$P(\text{LLW}) = Q_1 \cdot Q_2 \cdot P_3 = 0.8 \cdot 0.7 \cdot 0.9 = 0.504$

Total = $0.014 + 0.024 + 0.504 = 0.542$

Correct Options: (A), (C), (D)