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(B) Let $A$ be the event that one card is an ace of hearts and the other is not.There are two mutually exclusive cases:$(i)$ The first card is an ace

Vedclass · Accepted Answer

$\frac{1}{26}$

Vedclass · Answer

(B) Let $A$ be the event that one card is an ace of hearts and the other is not.There are two mutually exclusive cases:$(i)$ The first card is an ace of hearts and the second is not: $P(E_1) = \frac{1}{52} 	imes \frac{51}{51} = \frac{1}{52}$.$(ii)$ The first card is not an ace of hearts and the second is an ace of hearts: $P(E_2) = \frac{51}{52} 	imes \frac{1}{51} = \frac{1}{52}$.Therefore,the required probability is $P(E_1) + P(E_2) = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26}$.

Two cards are drawn without replacement from a well-shuffled pack. Find the probability that one of them is an ace of hearts.

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