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Question

(A) Let $A$ be the event that the card drawn from the first pack is an ace of hearts,and $B$ be the event that the card drawn from the second pack is

Vedclass · Accepted Answer

$\frac{103}{2704}$

Vedclass · Answer

(A) Let $A$ be the event that the card drawn from the first pack is an ace of hearts,and $B$ be the event that the card drawn from the second pack is an ace of hearts. The probability of drawing an ace of hearts from one pack is $P(A) = \frac{1}{52}$. The probability of $NOT$ drawing an ace of hearts from one pack is $P(A') = 1 - \frac{1}{52} = \frac{51}{52}$. The probability that neither card is an ace of hearts is $P(A' \cap B') = P(A') 	imes P(B') = \frac{51}{52} 	imes \frac{51}{52} = \frac{2601}{2704}$. The probability that at least one of them is an ace of hearts is $1 - P(A' \cap B') = 1 - \frac{2601}{2704} = \frac{2704 - 2601}{2704} = \frac{103}{2704}$.

One card is drawn from each of two ordinary packs of $52$ cards. The probability that at least one of them is an ace of hearts is:

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