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A box contains 24 identical bals of which 12 are white and remaining black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the 4th time on the 7th draw is

$\begin {array} {1 1} (A)\;\large\frac{5}{64} & \quad (B)\;\large\frac{27}{32} \\ (C)\;\large\frac{5}{32} & \quad (D)\;\large\frac{1}{2} \end {array}$


1 Answer

To satisfy the condition of the problem, first 6 draws should give exactly
3 white balls and 7th draw should give a white ball.
Hence the required probability is $ = 6C_3 \bigg(\large\frac{1}{2}\bigg)^3 \bigg(\large\frac{1}{2}\bigg)^3 \times \large\frac{1}{2}$
$ = \large\frac{20}{128}$
$ = \large\frac{5}{32}$
Ans : (C)


answered Jan 8, 2014 by thanvigandhi_1

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