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 box A, B, C respectively and A be the event that a screw selected at random is defective. Then, \[P({{E}_{1}})=P({{E}_{2}})=P({{E}_{3}})=\frac{1}{3}\] \[P\left( \frac{A}{{{E}_{1}}} \right)=\frac{1}{5},P\left( \frac{A}{{{E}_{2}}} \right)=\frac{1}{6},P\left( \frac{A}{{{E}_{3}}} \right)=\frac{1}{7}\] Using Baye's rule, Required probability \[=P\left( \frac{{{E}_{1}}}{A} \right)\] \[=\frac{P({{E}_{1}})P(A/{{E}_{1}})}{\left[ \begin{align} & P({{E}_{1}})P(A/{{E}_{1}})+P({{E}_{2}})P(A/{{E}_{2}}) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+P({{E}_{3}})P(A/{{E}_{3}}) \\ \end{align} \right]}\] \[=\frac{\frac{1}{3}\times \frac{1}{5}}{\frac{1}{3}\times \frac{1}{5}+\frac{1}{3}\times \frac{1}{6}+\frac{1}{3}\times \frac{1}{7}}=\frac{42}{107}\]
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