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(B) Total number of cards in a pack is $52$. There is only $1$ ace of heart in the pack.Let $A$ be the event that the first card is an ace of heart an

Vedclass · Accepted Answer

$\frac{1}{26}$

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(B) Total number of cards in a pack is $52$. There is only $1$ ace of heart in the pack.Let $A$ be the event that the first card is an ace of heart and $B$ be the event that the second card is an ace of heart.There are two mutually exclusive cases:$(i)$ The first card is an ace of heart and the second card is not an ace of heart:$P(E_1) = \frac{1}{52} 	imes \frac{51}{51} = \frac{1}{52}$$(ii)$ The first card is not an ace of heart and the second card is an ace of heart:$P(E_2) = \frac{51}{52} 	imes \frac{1}{51} = \frac{1}{52}$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 heart.

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