Sum

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`

Concept: Conditional Probability

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