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An electric component manufactured by 'RASU electronics' is tested for its defectiveness by a sophisticated testing device. Let $\mathrm{A}$ denote the event "the device is defective" and $\mathrm{B}$ the event "the testing device reveals the component to be defective". Suppose $\mathrm{P(A)=\alpha \quad and P(B / A)=P\left(B^{\prime} / A^{\prime}\right)=1-\alpha}$ , where $\mathrm{0<\alpha<1}$ . If the probability that the component is not defective, given that the testing device reveals it to be defective, is $\lambda$, then the value of $\mathrm{2008 %}$ must be
Option: 1
1004

Option: 2
1400

Option: 3
1040

Option: 4
$\frac{1}{2}$

Answers (1)

best_answer

Given that $\mathrm{P(A)=\alpha, P(B / A)=P\left(B^{\prime} / A^{\prime}\right)=1-\alpha}$
Thus,

$\mathrm{P\left(A^{\prime}\right) =1-P(A) }$

$\mathrm{=1-\alpha }$

and $\mathrm{P\left(B / A^{\prime}\right)=1-P\left(B^{\prime} / A^{\prime}\right)}$

$\mathrm{=1-(1-\alpha)}$

$\mathrm{=\alpha}$ ............(i)

$\mathrm{\therefore \quad P\left(A^{\prime} / B\right) =\frac{P\left(A^{\prime} \cap B\right)}{P(B)} }$

$\mathrm{ =\frac{P(B)-P(A \cap B)}{P(B)} }$

$\mathrm{ =\frac{P(B)-P(A) P(B / A)}{P(B)}\left\{\because P(B / A)=\frac{P(A \cap B)}{P(A)}\right\} }$

$\mathrm{ =\frac{P(B)-\alpha(1-\alpha)}{P(B)} }$ .................(ii)

But $\mathrm{ P(B) =P(A) \cdot P(B / A)+P\left(A^{\prime}\right) \cdot P\left(B / A^{\prime}\right) }$

$\mathrm{ =\alpha \cdot(1-\alpha)+(1-\alpha) \cdot \alpha }$ {from Eq.(i)}

$\mathrm{=2 \alpha(1-\alpha)}$

Putting the value of $\mathrm{P(B)}$ from Eq. (iii) in (ii), then

$\mathrm{P\left(A^{\prime} / B\right) =\frac{2 \alpha(1-\alpha)-\alpha(1-\alpha)}{2 \alpha(1-\alpha)} }$

$\mathrm{ =\frac{\alpha(1-\alpha)}{2 \alpha(1-\alpha)} }$

$\mathrm{ =\frac{1}{2}=\lambda \text { (given) } }$

$\mathrm{ \therefore \quad 2008 \lambda =2008 \times \frac{1}{2} }$

$=1004$

Hence option 1 is correct.

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manish

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