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 as the captain, the probability that the team wins the tournament is $0.8$ and if $B$ is selected as the captain, the probability that the team wins the tournament is $0.7$. Then the probability, that the team wins the tournament, is:
- A$0.74$
- B$0.76$
- C$0.72$
- D$0.78$
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If $A$ and $B$ are two events,then which of the following is $NOT$ true?
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View SolutionIn a game, two dice are thrown simultaneously by a person $A$ and two cards are drawn at random simultaneously from a pack of $52$ playing cards by a person $B$. They win the game if $A$ gets a prime score as the sum of the numbers appearing on both the dice and $B$ gets a face card and a card having a prime number. Then the probability that both $A$ and $B$ win is:
All face cards from a pack of $52$ playing cards are removed. From the remaining $40$ cards,two are drawn randomly without replacement. The probability of drawing a pair (cards of the same denomination) is:
Three dice are thrown simultaneously and the sum of the numbers appeared on them is noted. If $A$ is the event of getting a sum greater than $14$ and $B$ is the event of getting a sum which is a multiple of $3$,then $P(A \cap \overline{B}) + P(\overline{A} \cap B) = $
For independent events $A$ and $B$,if $P(A) = \frac{1}{2}$ and $P(A \cup B) = \frac{3}{5}$,then $P(B) =$ . . . . . . .
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