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A box contains 2 black,4 white, and 3 red balls. One ball is drawn at random from the box and kept aside. From
the remaining balls in the box, another ball is drawn at random and kept aside the first. This process is repeated
till all the balls are drawn from the box. The probability that the balls drawn are in the sequence of 2 black, 4 white, and 3 red is

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1/1260

1/7560

$\frac{1}{26}$

None of these

answer is A.

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Detailed Solution

The required probability is $\frac{2}{9} \times \frac{1}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times 1 \times 1 \times 1 = \frac{1}{1260}$

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