Question

A and B are two independent events of a random experiment and \(P(A)>P(B)\). If the probability that both A and B occur is \(\frac{1}{6}\) and neither of them occurs is \(\frac{1}{3}\), then the probability of the occurrence of B is:

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{3}{8}\)

Hint:

Use independence and union formulas to set up equations, then solve quadratic equations carefully considering inequalities.

Answer 1

The Correct Option is B

Given \(P(A \cap B) = \frac{1}{6}\) and \(P(\text{neither A nor B}) = \frac{1}{3}\). Since A and B are independent, \[ P(A \cap B) = P(A) P(B) = \frac{1}{6}. \] Also, \[ P(\text{neither A nor B}) = 1 - P(A \cup B) = \frac{1}{3} \implies P(A \cup B) = \frac{2}{3}. \] Using the formula, \[ P(A \cup B) = P(A) + P(B) - P(A)P(B) = \frac{2}{3}. \] Let \(P(B) = x\) and \(P(A) = \frac{1}{6x}\) (since \(P(A)P(B) = \frac{1}{6}\)). Substitute: \[ \frac{1}{6x} + x - \frac{1}{6} = \frac{2}{3}. \] Multiply both sides by \(6x\): \[ 1 + 6x^2 - x = 4x, \] \[ 6x^2 - 5x + 1 = 0. \] Solve quadratic: \[ x = \frac{5 \pm \sqrt{25 - 24}}{12} = \frac{5 \pm 1}{12}. \] Possible values: \[ x = \frac{6}{12} = \frac{1}{2}
\text{or}
x = \frac{4}{12} = \frac{1}{3}. \] Since \(P(A)>P(B)\), \(P(B) = \frac{1}{3}\).