Question

A computer producing factory has only two plants T₁ and T₂ produces 20% and plant T₂ produces 80% of the total computers produced. 7% of the 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 T₁ 10P(computer turns out to be defective given that it is produced in plant T₂ ). 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 T₂ is

Answer 1

Given: \(P(T_1)=0.2,\ P(T_2)=0.8,\ P(D)=0.07,\) and \(P(D\mid T_1)=10\,P(D\mid T_2)\).

Let \(p_2=P(D\mid T_2)\). Then \(P(D\mid T_1)=10p_2\). Using total probability: \[ 0.07=P(D)=0.2(10p_2)+0.8(p_2)=(2+0.8)p_2=2.8p_2 \Rightarrow p_2=\frac{0.07}{2.8}=0.025. \] Hence \(P(D\mid T_1)=0.25\).

We need: \(P(T_2\mid \overline D)=\dfrac{P(T_2)\,P(\overline D\mid T_2)}{P(\overline D)}\), where \(P(\overline D)=1-0.07=0.93\) and \(P(\overline D\mid T_2)=1-0.025=0.975\).

\[ P(T_2\mid \overline D)=\frac{0.8\times 0.975}{0.93} =\frac{0.78}{0.93} =\frac{26}{31}\approx 0.8387. \]

Answer: \(\boxed{\dfrac{26}{31}}\).

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