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? A. pq + (1 – p) (1 – q) B. (1 – q)p C. (1 – p)q D. pq

The correct answer is $\boxed{pq}$.

Let $A$ be the event that a computer is faulty and $B$ be the event that a computer is declared faulty. We are given that $P(A) = p$ and $P(B|A) = q$. We want to find $P(A \cap B)$.

By the definition of conditional probability, $P(B|A) = \frac{P(A \cap B)}{P(A)}$. Substituting in the given values, we get $q = \frac{P(A \cap B)}{p}$. Multiplying both sides by $p$, we get $pq = P(A \cap B)$.

Therefore, the probability of a computer being declared faulty is $pq$.

Option A is incorrect because it includes the term $(1 – p) (1 – q)$. This term represents the probability that a computer is not faulty and the testing process gives a correct result. This is not the event that we are interested in.

Option B is incorrect because it is the probability that a computer is not faulty. This is not the event that we are interested in.

Option C is incorrect because it is the probability that a computer is faulty and the testing process gives an incorrect result. This is not the event that we are interested in.