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

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 heart,and $B$ be the event that the card drawn from the second pack is an ace of heart.The probability of drawing an ace of heart from a pack of $52$ cards is $P(A) = P(B) = \frac{1}{52}$.The probability of $NOT$ drawing an ace of heart from a pack is $P(A') = P(B') = 1 - \frac{1}{52} = \frac{51}{52}$.The probability that at least one of them is an ace of heart is given by $1 - P(	ext{none is an ace of heart})$.Since the draws are independent,$P(	ext{none is an ace of heart}) = P(A') 	imes P(B') = \frac{51}{52} 	imes \frac{51}{52} = \frac{2601}{2704}$.Therefore,the required probability is $1 - \frac{2601}{2704} = \frac{2704 - 2601}{2704} = \frac{103}{2704}$.

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 heart,is

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