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Consider a company that assembles computers. The probability of a faulty assembly of any computer is $p$. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of $q$. What is the probability of a computer being declared faulty?

1. $pq + (1 - p)(1 - q)$
2. $(1 - q)p$
3. $(1 - q)p$
4. $pq$
edited | 1.7k views

in image below the ticks shows those branch where the result is declared as faulty.

so required probability = sum of those two branches $= pq + (1-p)(1-q)$

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nice amar
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Can we say here that

A : Assembly is faulty

B : Test gives correct result

A and B are independent

Hence P(A) = P(A $\cap$ B) + P(A $\cap$ B') = [P(A)*P(B) + P(A)*P(B')]
I think a is the correct option ..

prob = pq + (1-p)(1-q)  // in words it means

(it is faulty)(machine detects correctly) + (it is not faulty)(machine detects incorrectly)
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missed + between pq and (1 - p)(1 -q)

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