Q33 / 75 · JEE Main 2026 (05 April Shift 2) · Mathematics
The probabilities that players $A$ and $B$ of a team are selected for the captaincy for a tournament are $0.6$ and $0.4$, respectively. If $A$ is selected the captain, the probability that the team wins the tournament is $0.8$ and if $B$ is selected the captain, the probability that the team wins the tournament is $0.7$. Then the probability, that the team wins the tournament, is :
- A$0.74$
- Correct answerB$0.76$
- C$0.72$
- D$0.78$
Solution
Let $A$ be the event that player A is selected as captain, and $B$ the event that player B is selected as captain. Let $W$ be the event that the team wins the tournament.
Given:
$P(A) = 0.6$, $P(B) = 0.4$
$P(W \mid A) = 0.8$, $P(W \mid B) = 0.7$
By the law of total probability:
$P(W) = P(A) \cdot P(W \mid A) + P(B) \cdot P(W \mid B)$
$P(W) = (0.6)(0.8) + (0.4)(0.7)$
$P(W) = 0.48 + 0.28 = 0.76$
Hence, the correct option is $(2)\ 0.76$.
Given:
$P(A) = 0.6$, $P(B) = 0.4$
$P(W \mid A) = 0.8$, $P(W \mid B) = 0.7$
By the law of total probability:
$P(W) = P(A) \cdot P(W \mid A) + P(B) \cdot P(W \mid B)$
$P(W) = (0.6)(0.8) + (0.4)(0.7)$
$P(W) = 0.48 + 0.28 = 0.76$
Hence, the correct option is $(2)\ 0.76$.
Hence, the correct answer is option: 'B'.