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+11 votes

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. $pq$
asked in Probability by Boss (18.3k points)
edited by | 1.7k views

2 Answers

+26 votes
Best answer

answer = option A

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

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

answered by Boss (30.9k points)
edited by
nice amar
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')]
+13 votes
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)
answered by Boss (14.4k points)
edited by
missed + between pq and (1 - p)(1 -q)

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