NIMCET 2019 Question Paper with solutions

Question : 44 of 120

Marks: +1, -0

A computer producing factory has only two plaats

\T_1

and

\T_2

- Plant

\T_1

produces 20% and plant T2 produces 80% of the total computer produced. 7% of the computers produced in the factory turn out to be defective. It is known that P(computer turns out to the 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

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

\T_2

$36/73$
$47/79$
$78/93$
$75/83$

Validate

Solution:

Let

T_1

and

T_2

be the event that computer is produced by plant

T_1

and

T_2

respectively. Again, let D and ND be the event that computer produced is defective and non-defective respectively

P(T_1)=20/100,P(T_2)=80/100,P(D)=7/100

Again let

P(D/T_1)=10\;\;P(D/T_2)=k

∴\;\;P(D/T_1)=k

and

P(D/T_2)=k

Now,

P(D)=P(T_1)P(D/T_1)+P(T_2)P(D/T_2)

⇒7/100=20/100×k+80/100×k/10

⇒7=20k+8k

⇒k=1/4

Now, required probability

=P(T_2{/}{ND})

=\;\;{P(T_{z})P(ND{/}T_{2})}/{P(T_{1})P(ND{/}T_{1})+P(T_{2})P(ND{/}T_{2})}

=\;\;{\;\;{80}/{100}×(1-\;\;{1}/{40})}/{\;\;{20}/{100}(1-\;\;{1}/{4})+\;\;{80}/{100}(1-\;\;{1}/{40})}=\;\;{4×\;\;{39}/{40}}/{{3}/{4}+4×{39}/{40}}

={4×39}/{30+4×39}={156}/{186}={78}/{93}

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