Question:
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 :
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 :
Updated On: Apr 12, 2026
- 0.74
- 0.76
- 0.72
- 0.78
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
This problem uses the Law of Total Probability. The event "winning" is partitioned based on who the captain is.
: Key Formula or Approach:
$P(W) = P(A)P(W|A) + P(B)P(W|B)$.
Step 2: Detailed Explanation:
$P(A) = 0.6$ (Player A is captain).
$P(B) = 0.4$ (Player B is captain).
$P(W|A) = 0.8$ (Winning probability given A is captain).
$P(W|B) = 0.7$ (Winning probability given B is captain).
Total winning probability:
$P(W) = (0.6)(0.8) + (0.4)(0.7)$.
$P(W) = 0.48 + 0.28 = 0.76$.
Step 3: Final Answer:
The probability of winning is 0.76.
This problem uses the Law of Total Probability. The event "winning" is partitioned based on who the captain is.
: Key Formula or Approach:
$P(W) = P(A)P(W|A) + P(B)P(W|B)$.
Step 2: Detailed Explanation:
$P(A) = 0.6$ (Player A is captain).
$P(B) = 0.4$ (Player B is captain).
$P(W|A) = 0.8$ (Winning probability given A is captain).
$P(W|B) = 0.7$ (Winning probability given B is captain).
Total winning probability:
$P(W) = (0.6)(0.8) + (0.4)(0.7)$.
$P(W) = 0.48 + 0.28 = 0.76$.
Step 3: Final Answer:
The probability of winning is 0.76.
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