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Option 3 : Quantity I ˃ Quantity II
Quantity I:
Total number of outcomes = 20C4
= 20!/(4! × 16!)
= (20 × 19 × 18 × 17 × 16!) / (4 × 3 × 2 × 1 × 16!)
= 4845
Number of favourable outcomes = 6C2 × 4C2
= 90
⇒ Required probability = 90/4845
= 6/323
= 0.0186 (approx)
Quantity II:
Total number of possible ways to drawn a card = 52C5
= 52!/(5! × 47!)
= (52 × 51 × 50 × 49 × 48 × 47!) / (5 × 4 × 3 × 2 × 1 × 47!)
= 2598960
Number of sets of 5 with all are jack = 4C4 × 48C1
= 1 (48 × 47!) / (1! × 47!) = 48
⇒ P(E) = 48/2598960
= 1/54145
= 0.000018(approx)
∴ Quantity I > Quantity II
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