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Question
From a pack of well-shuffled cards, two cards are drawn at random. Find the probability that both the cards are diamonds when first card drawn is kept aside
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Solution
In a pack of 52 cards, there are 13 diamond cards.
Let event A: The first card drawn is a diamond card.
∴ P(A) = `(""^13"C"_1)/(""^52"C"_1)`
= `13/52`
= `1/4`
Let event B: The second card drawn is a diamond card.
Since the first diamond card is kept aside, we now have 51 cards, out of which 12 are diamond cards.
∴ Probability that the second card is a diamond card under the condition that the first diamond card is kept aside in the pack
= `"P"("B"//"A")`
= `(""^12"C"_1)/(""^52"C"_1)`
= `12/51`
= `4/17`
∴ Required probability = P(A ∩ B)
= `"P"("B"//"A") * "P"("A")`
= `1/4 xx 4/17`
= `1/17`
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