If 3 balls are drawn at random, what is the probability that 1 is green and the other 2 are black ?

\frac{2}{85}

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\frac{4}{87}

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\frac{1}{85}

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\frac{3}{85}

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\frac{5}{85}

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Solution

The correct option is D $\frac{3}{85}$
$^{17} C_{3} = 17 \times 16 \times \frac{15}{3} \times 2 \times 1 = \frac{4080}{6} = 680^{4} C_{1} \times^{4} C_{2} = 4 \times 4 \times \frac{3}{2} \times 1 = 24 P = \frac{24}{680} = \frac{3}{85}$

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If 3 balls are drawn at random, what is the probability that 1 is green and the other 2 are black ?

The correct option is D 38517C3=17×16×153×2×1=40806=6804C1×4C2=4×4×32×1=24P=24680=385

The correct option is D $\frac{3}{85}$
$^{17} C_{3} = 17 \times 16 \times \frac{15}{3} \times 2 \times 1 = \frac{4080}{6} = 680^{4} C_{1} \times^{4} C_{2} = 4 \times 4 \times \frac{3}{2} \times 1 = 24 P = \frac{24}{680} = \frac{3}{85}$