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Suppose that chips for an integrated circuit are tested and that the probability

that they are detected if they are defective is $0.95$, and the probability that

they are

declared sound if in fact they are sound is $0.97$. If .5% of the chips are

faulty, what

is the probability that a chip that is declared faulty is sound?

What i tried

Let $A_1$ be the event that the chips are faulty

Let $A_2$ be the event that the chips are sound

Let $B$ be the event that the chips are declared faulty

Using Baytes formula for conditional Probaility

$$P(A_{2}/B)=\frac{P(B/A_{2}).P(A_{2})}{P(B/A_{2}).P(A_{2})+P(B/A_{1}).P(A_{1})}$$

Plugging in the values to the formula

I got $$P(A_{2}/B)=\frac{(0.05).(0.995)}{(0.05).(0.995)+(0.95).(\frac{0.5}{100})}$$

Howver i dont seem to get the right answer. I know the right answer is 0.863. Could anyone explain this question. Thanks

Ethan Bolker
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ys wong
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1 Answers1

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I guess you used the wrong formula $P(B|A_2)=1-P(B|A_1)$, where you should have used $P(B|A_2)=1-P(\bar{B}|A_2)=1-0.97$.

Augustin
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