The chances of a defective screw in three boxes A, B, C are \(\frac{1}{5},{\rm{\;}}\frac{1}{6}\) and \(\frac{1}{7}\) respectively. A box is selected at random and a screw drawn from it at random is found to be defective. Find the probability that it came from box A.

Question

Question

Answer 1

Let E₁, E₂ and E₃ 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(E₁) = P(E₂) = P(E₃) = 1/3,

\({\rm{P}}\left( {{\rm{A}}/{{\rm{E}}_1}} \right) = \frac{1}{5}\)

\({\rm{P}}\left( {\frac{{\rm{A}}}{{{{\rm{E}}_2}}}} \right) = \frac{1}{6} \Rightarrow {\rm{P}}\left( {{\rm{A}}/{{\rm{E}}_3}} \right) = \frac{1}{7}\)

Then, by Baye’s theorem, required probability

= P(E₁/A)

\(= \frac{{\frac{1}{3}.\frac{1}{5}}}{{\frac{1}{3}.\frac{1}{5} + \frac{1}{3}.\frac{1}{6} + \frac{1}{3}.\frac{1}{7}}} = \frac{{42}}{{107}}\)