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Q.

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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a

$\frac{2601}{2704}$

b

$\frac{103}{2704}$

c

$\frac{1}{2704}$

d

$\frac{2}{52}$

answer is A.

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Detailed Solution

Required probability is 1-P (no ace of heart)
$= 1 - \frac{51}{52} \cdot \frac{51}{52} = \frac{(52 + 51)}{52.52} = \frac{103}{2704}$

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