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There are 8 brown balls, 4 orange balls and 5 black balls in a bag. Five balls are chosen at random. What is the probability of their being 2 brown balls, 1 orange ball and 2 black balls?

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$\frac{191}{1547}$

$\frac{180}{1547}$

$\frac{280}{1547}$

$\frac{189}{1547}$

answer is C.

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Detailed Solution

Total possible outcomes $=^{17} C_{5}$

$= \frac{17 \times 16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4 \times 5} = 6188$

Total favorable outcomes $=^{8} C_{2} \times^{4} C_{1} \times^{5} C_{2}$

$= \frac{8 \times 7}{1 \times 2} \times 4 \times \frac{5 \times 4}{1 \times 2} = 28 \times 4 \times 10 = 1120$

Required probability

$= \frac{1120}{6188} = \frac{280}{1547}$

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