JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Probability XII

A computer producing factory has only two plants

T_{1}

and

T_{2}

. Plant

T_{1}

produces 20% and plant

T_{2}

produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that

P

(computer turns out to be defective given that it is produced in plant

T_{1}

)

= 10 P

(computer turns out to be defective given that it is produced in plant

T_{2}

),
where

P (E)

denotes the probability of an event

E

. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant

T_{2}

Answer: (C)

Solution

Let

E_{1}

: computer from plant

T_{1}

E_{2}

: computer from plant

T_{2}

A

: computer is defective.

Given:

P (E_{1}) = 0.2

P (E_{2}) = 0.8

P (A) = 0.07

, and

P (A ∣ E_{1}) = 10 P (A ∣ E_{2})

.

Using total probability:

P (A) = P (A ∣ E_{1}) P (E_{1}) + P (A ∣ E_{2}) P (E_{2})

0.07 = P (A ∣ E_{1}) \cdot 0.2 + P (A ∣ E_{2}) \cdot 0.8

Substituting

P (A ∣ E_{1}) = 10 P (A ∣ E_{2})

0.07 = 10 P (A ∣ E_{2}) \cdot 0.2 + P (A ∣ E_{2}) \cdot 0.8 = 2.8 P (A ∣ E_{2})

P (A ∣ E_{2}) = \frac{0.07}{2.8} = 0.025

P (A ∣ E_{1}) = 0.25

.

Now,

P (\overline{A} ∣ E_{2}) = 1 - 0.025 = 0.975

and

P (\overline{A}) = 1 - 0.07 = 0.93

.

Using Bayes' theorem:

P (E_{2} ∣ \overline{A}) = \frac{P ( A ∣ E _{2} ) P ( E _{2} )}{P ( A )} = \frac{0.975 \times 0.8}{0.93} = \frac{0.78}{0.93} = \frac{78}{93}

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Single correct option