There are 6 blue pens, 5 black pens and 4 green pens in a bag. Three pens are chosen randomly.
Quantity I – The probability of their being 2 blue and 1 green pens.
Quantity II – The probability of their being 1 blue, 1 black and 1 green pens.
Total number of pens = 6 + 5 + 4 = 15 Three pens are chosen at random. Then, n(S) = ¹⁵C₃ = (15 ×14 ×13 )/(3 ×2 ×1) = 130 Quantity I. n(E) = ⁶C₂ × ⁴C₁ = 15 × 4 = 60 P(E) = 60/130 = 6/13 Quantity II. n(E) = ⁶C₁ × ⁵C₁ × ⁴C₁ = 6 × 5 × 4 = 120 P(E) = 120/130 = 12/13 ∴ Quantity I < Quantity II.
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