QJEEAdv16Ma6d1b: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Ma6d1b Solution: A computer producing factory has only two plants and Plant…

Question

Verified Solution • 2025–26 Session moderate

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}

)

= 10P

(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}

is

Accepted Answer

Let

P_1

be the defective computers that are produced from plant

T_1

and

P_2

be that from plant

T_2

.

The total percentage of the defective computers produced is

7 \%

.

Now,

P_1=10 P

and

P_2=P

.

The computers produced that are defective:

\begin{gathered}\frac{20}{100} \times P_1+\frac{80}{100} \times P_2=\frac{7}{100} \\\\20 P_1+80 P_2=7 \\\\200 P+80 P=7 \\\\P=\frac{7}{280}=\frac{1}{40}\end{gathered}

Now, the probability of the defective products is calculated as follows:

\frac{20}{100} P_1+\frac{80}{100} P_2=\frac{20}{100} \times \frac{1}{4}+\frac{80}{100} \times \frac{1}{40}=\frac{28}{400}=\frac{7}{100}

The probability of producing NOT defective computers is

1-\frac{7}{100}=\frac{93}{100}

The probability that plant

T_2

produces NOT defective computers is calculated as follows:

\frac{(80 / 100) \times(39 / 40)}{(93 / 100)}=\frac{78}{93}

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