A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{4}\]
D) \[\frac{3}{4}\]
Correct Answer: C
Solution :
Let x be the probability of success in each trial, then \[(1-x)\] will be the probability of failure in each trial. Thus, probability of exactly two successes in a series of trials \[=P({{\bar{E}}_{1}}{{E}_{2}}{{E}_{3}}+{{E}_{1}}{{\bar{E}}_{2}}{{E}_{3}}+{{E}_{1}}{{E}_{2}}{{\bar{E}}_{3}})\] \[=(1-x)x.x+x(1-x)x+x.x(1-x)\] \[=3{{x}^{2}}(1-x)\] and the probability of three successes \[P({{E}_{1}}{{E}_{2}}{{E}_{3}})=x.x.x={{x}^{3}}\] According to question \[9{{x}^{3}}=3{{x}^{2}}(1-x)\] \[\Rightarrow \] \[3x=1-x\] \[\Rightarrow \] \[4x=1\] \[\Rightarrow \] \[x=\frac{1}{4}.\] Hence, the probability of success in each trial is\[\frac{1}{4}.\]
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