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 :

Let the events be defined as follows:
  • $$C_A$$ : A is chosen as captain, with $$P(C_A)=0.6$$.
  • $$C_B$$ : B is chosen as captain, with $$P(C_B)=0.4$$.
  • $$W$$ : team wins the match.

The conditional probabilities of winning are given by:
  • $$P(W\,\lvert\,C_A)=0.8$$ (win if A is captain)
  • $$P(W\,\lvert\,C_B)=0.7$$ (win if B is captain)

Because exactly one of A or B is made captain, the events $$C_A$$ and $$C_B$$ form a partition of the sample space. Therefore, by the Law of Total Probability, the overall probability of winning is

$$\begin{aligned} P(W) &= P(W\,\lvert\,C_A)\,P(C_A) + P(W\,\lvert\,C_B)\,P(C_B)\\ &= 0.8 \times 0.6 \;+\; 0.7 \times 0.4\\ &= 0.48 + 0.28\\ &= 0.76. \end{aligned}$$

Hence, the required probability of winning is $$0.76$$.

Option B which is: $$0.76$$