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VITEEE 2010 Paper

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Question : 92 of 120

Marks: +1, -0

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

$\frac{16}{29}$
$\frac{1}{15}$
$\frac{27}{59}$
$\frac{42}{107}$

Solution:

Let

E_{1},E_{2}\text{ and }E_{3}

denote the events of selecting boxes A, B, C respectively and A be the event that a screw selected 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},

By Baye's rule, the required probability

P(E_{1}/A)=\frac{P(E_{1})P(A/E_{1})}{P(E_{1})P(A/E_{1})+P(E_{2})P(A/E_{2})+P(E_{3})P(A/E_{3})}

⇒

P(E_{1}/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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