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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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