A) \[\frac{16}{29}\]
B) \[\frac{1}{15}\]
C) \[\frac{27}{59}\]
D) \[\frac{42}{107}\]
Correct Answer: D
Solution :
Let \[{{E}_{1}},\,{{E}_{2}}\]and \[{{E}_{3}}\]denote the events of selecting boxes \[A,\]\[B,\]\[C\]respectively at random is defective. Then, \[P({{E}_{1}})=\frac{1}{3},P({{E}_{2}})=\frac{1}{3},P({{E}_{3}})=\frac{1}{3}\] \[P(A/{{E}_{1}})=\frac{1}{5},P(A/{{E}_{2}})=\frac{1}{6},P(A/{{E}_{3}})=\frac{1}{7}\] Now, by Bayes rule, the required, probability \[P({{E}_{1}}\text{/}A)=\]\[\frac{P({{E}_{1}})P(A\text{/}{{E}_{1}})}{P({{E}_{1}})P(A\text{/}{{E}_{1}})+P({{E}_{2}})P(A\text{/}{{E}_{2}})+P({{E}_{3}})P(A\text{/}{{E}_{3}})}\] \[\Rightarrow \]\[P({{E}_{1}}\text{/}A)=\frac{\frac{1}{3}\cdot \frac{1}{5}}{\frac{1}{3}\cdot \frac{1}{5}+\frac{1}{3}\cdot \frac{1}{6}+\frac{1}{3}\cdot \frac{1}{7}}\] \[=\frac{\frac{1}{5}}{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}}\] \[=\frac{42}{107}\]
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