Question

The chances of defective screws in three boxes A, B, C are $\frac { 1 } { 5 } , \frac { 1 } { 6 } , \frac { 1 } { 7 }$ respectively. A box is selected at random and a

screw drawn from it at random is found to be defective. Then the probability that it came from box A is

• A. 16/29
• B. 1/15
• C. 27/59
• D. 42/107

Answer 1

Answer: (D)

42/107

Answer 2

Let $\mathrm { E } _ { 1 } , \mathrm { E } _ { 2 }$ and denote the events of selecting box A,B,C respectively and A be the event that a screw selected at random is defective.

Then $\mathrm { P } \left( \mathrm { E } _ { 1 } \right) = \mathrm { P } \left( \mathrm { E } _ { 2 } \right) = \mathrm { P } \left( \mathrm { E } _ { 3 } \right) = 1 / 3$

$\mathrm { P } \left( \mathrm { A } / \mathrm { E } _ { 1 } \right) = \frac { 1 } { 5 } , \mathrm { P } \left( \mathrm { A } / \mathrm { E } _ { 2 } \right) = \frac { 1 } { 6 } , \mathrm { P } \left( \mathrm { A } / \mathrm { E } _ { 3 } \right) = \frac { 1 } { 7 }$

By Baye's rule,

Required probability $\mathrm { P } \left( \mathrm { E } _ { 1 } / \mathrm { A } \right)$

$= \frac { P \left( E _ { 1 } \right) \mathrm { P } \left( \mathrm { A } / \mathrm { E } _ { 1 } \right) } { \mathrm { P } \left( \mathrm { E } _ { 1 } \right) \mathrm { P } \left( \mathrm { A } / \mathrm { E } _ { 1 } \right) + \mathrm { P } \left( \mathrm { E } _ { 2 } \right) \mathrm { P } \left( \mathrm { A } / \mathrm { E } _ { 2 } \right) + \mathrm { P } \left( \mathrm { E } _ { 3 } \right) \mathrm { P } \left( \mathrm { A } / \mathrm { E } _ { 3 } \right) }$ $= \frac { \frac { 1 } { 3 } \times \frac { 1 } { 5 } } { \frac { 1 } { 3 } \times \frac { 1 } { 5 } \times \frac { 1 } { 3 } \times \frac { 1 } { 6 } \times \frac { 1 } { 3 } \times \frac { 1 } { 7 } } = \frac { 42 } { 107 }$