A) \[\frac{5}{32}\]
B) \[\frac{27}{32}\]
C) \[\frac{23}{32}\]
D) \[\frac{9}{32}\]
Correct Answer: B
Solution :
[b] Let the second success occur at the nth trial. This means that there was exactly one success in the first n - 1 trials, so that the probability of getting the second success at the nth trial is \[{{p}_{n}}={{(}^{n-1}}{{C}_{1}}p{{q}^{n-1-1}})p=(n-1){{p}^{2}}{{q}^{n-2}}\] Therefore the probability of the required event is \[{{p}_{4}}+{{p}_{5}}+{{p}_{6}}+...=3{{p}^{2}}{{q}^{2}}+4{{p}^{2}}{{q}^{3}}+5{{p}^{2}}{{q}^{4}}+6{{p}^{2}}{{q}^{5}}+...\]\[={{p}^{2}}{{q}^{2}}(3+4q+5{{q}^{2}}+6{{q}^{3}}+...)\] \[={{p}^{2}}{{q}^{2}}[3(1+q+{{q}^{2}}+{{q}^{3}}+...)+q(1+2q+3{{q}^{2}}+...)]\]\[={{p}^{2}}{{q}^{2}}[(3{{(1-q)}^{-1}}+q{{(1-q)}^{-2}}]={{p}^{2}}{{q}^{2}}\left( \frac{3}{p}+\frac{q}{{{p}^{2}}} \right)\] \[={{q}^{2}}(3p+q)={{q}^{2}}(2p+p+q)={{q}^{2}}(2p+1)\] \[={{\left( \frac{3}{4} \right)}^{2}}\left[ 2\left( \frac{1}{4} \right)+1 \right]=\frac{9}{12}\left( \frac{1}{2}+1 \right)=\frac{27}{32}\]
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