If P(A) = 0.4 , P(B) = 0.8 and P(A|B) = 0.6 , then P(A ∪ B) is:

The correct answer is Option (2) → 0.72

By the multiplication rule:

    \(P(A \cap B) = P(B)\,P(A|B)\)
    \(= 0.8 \times 0.6 = 0.48\)

By the addition rule:

    \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
    \(= 0.4 + 0.8 - 0.48\)
    \(= 0.72\)

Note: It seems that there is a mistake in the original NTA paper: the given values do not fit together. They give P(A ∩ B) = 0.48, which is more than P(A) = 0.4, although A ∩ B is a part of A; in the same way the result 0.72 is less than P(B) = 0.8, although B is a part of A ∪ B. The question can still be solved, because the multiplication and addition rules applied to the given values lead to 0.72, which is Option (2).