A) \[\frac{1}{42}\]
B) \[\frac{5}{42}\]
C) \[\frac{1}{6}\]
D) \[\frac{2}{21}\]
Correct Answer: B
Solution :
[b] Since, four balls can be drawn alternatively in two ways - RWRW or WRWR. If red ball be drawn first then the probability of drawing balls alternatively \[=\frac{6}{9}\times \frac{3}{8}\times \frac{5}{7}\times \frac{2}{6}=\frac{5}{3\times 4\times 7}=\frac{5}{84}\] If white ball be drawn first then the probability of drawing balls alternatively \[=\frac{3}{9}\times \frac{6}{8}\times \frac{2}{7}\times \frac{5}{6}=\frac{5}{84}\] But both cases be mutually exclusive. So, the required probability \[=\frac{5}{84}+\frac{5}{84}=\frac{2\times 5}{84}=\frac{5}{42}\] Hence, option [b] is correct.
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