A king, queen and jack of diamonds are removed from a pack of 52 cards and then the pack is well-shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of (i) diamonds, (ii) a jack.

Solution

When a king, queen and jack of diamonds are removed from a pack of 52 cards, then possible outcome(s)are
52–3 = 49 i.e., n(S) = 49
I. Let A be the favourable outcomes of getting a card of diamonds. Then i.e., n( A) = 10
Therefore,
$rightwards double arrow space space space space space straight P left parenthesis straight A right parenthesis space equals space fraction numerator straight n left parenthesis straight A right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 10 over 49$

II. Let B be the favourable outcomes of getting a jack, then
i.e., n(B) = 3
Therefore,

$rightwards double arrow space space space space space space straight P left parenthesis straight B right parenthesis space equals space fraction numerator straight n left parenthesis straight B right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 3 over 49$

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Real Numbers

A king, queen and jack of diamonds are removed from a pack of 52 cards and then the pack is well-shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of (i) diamonds, (ii) a jack.

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