Let A and B be two events such that P(A) = 0.6 and P(B) = 0.7. The least possible value of P(A|B) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{3}{7}\)
By the definition of conditional probability,
\( \displaystyle P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{P(A\cap B)}{0.7} \)
so P(A|B) is least when \(P(A\cap B)\) is least. By the addition theorem, and since no probability exceeds 1,
\(P(A\cap B) = P(A)+P(B)-P(A\cup B)\)
\(P(A\cap B) \ge 0.6+0.7-1 = 0.3\)
The value 0.3 is reached when \(A\cup B\) is the whole sample space. So the least value is
\( \displaystyle P(A|B) = \frac{0.3}{0.7} = \frac{3}{7} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{10}\) — Incorrect. This is the least value of \(P(A\cap B)\), not yet divided by P(B).
Option (3) \(\displaystyle\frac{6}{7}\) — Incorrect. This is \(\frac{0.6}{0.7}\), the greatest possible value, reached when A lies inside B.
Option (4) \(\displaystyle\frac{3}{5}\) — Incorrect. This is P(A), the value only when A and B are independent; it is not the least.