A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{2}{3}\]
D) \[\frac{1}{4}\]
Correct Answer: D
Solution :
Let R stand for drawing red ball \[B\] for drawing black ball and \[W\] for drawing white ball. Then required probability \[=P(WWR)+P(BBR)+P(WBR)+P(BWR)+P(WRR)+\] \[P(BRR)+P(RWR)+P(RBR).\] \[=\frac{3.2.2}{8.7.6}+\frac{3.2.2}{8.7.6}+\frac{3.3.2}{8.7.6}+\frac{3.3.2}{8.7.6}+\frac{3.2.1}{8.7.6}\] \[+\frac{3.2.1}{8.7.6}+\frac{2.3.1}{8.7.6}+\frac{2.3.1}{8.7.6}\] \[=\frac{2}{56}+\frac{2}{56}+\frac{3}{56}+\frac{3}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}=\frac{1}{4}\].You need to login to perform this action.
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