Question

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

• A. \(0 \text{ or } 1\)
• B. \(\frac{1}{2} \text{ or } \frac{1}{4}\)
• C. \(\mathbf{\frac{1}{2} \text{ or } \frac{1}{3}}\)
• D. \(\frac{1}{2} \text{ or } \frac{1}{7}\)

Hint:

When two events are independent, use \(P(A \cap B) = P(A) \cdot P(B)\), and remember that the probability of neither event occurring is the product of their individual non-occurrence probabilities.

Answer 1

The Correct Option is C

Let \(P(A) = p\), \(P(B) = q\). Since \(A\) and \(B\) are independent, \[ P(A \cap B) = pq = \frac{1}{6} \tag{1} \] Also, the probability that neither occurs is: \[ P(A^c \cap B^c) = (1 - p)(1 - q) = \frac{1}{3} \tag{2} \] Let’s solve equations (1) and (2): From (1): \(q = \frac{1}{6p}\) Substitute in (2): \[ (1 - p)\left(1 - \frac{1}{6p}\right) = \frac{1}{3} \Rightarrow (1 - p)\left(\frac{6p - 1}{6p}\right) = \frac{1}{3} \Rightarrow \frac{(1 - p)(6p - 1)}{6p} = \frac{1}{3} \] Multiply both sides by \(6p\): \[ (1 - p)(6p - 1) = 2p \Rightarrow 6p - 1 - 6p^2 + p = 2p \Rightarrow -6p^2 + 5p - 1 = 2p \Rightarrow -6p^2 + 3p - 1 = 0 \Rightarrow 6p^2 - 3p + 1 = 0 \] Solving the quadratic: \[ p = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6} = \frac{3 \pm \sqrt{9 - 24}}{12} = \frac{3 \pm \sqrt{-15}}{12} \] Oops! This gives complex roots. Let's recheck our algebra. Rewriting equation (2): \[ (1 - p)(1 - \frac{1}{6p}) = \frac{1}{3} \Rightarrow 1 - \frac{1}{6p} - p + \frac{p}{6p} = \frac{1}{3} \Rightarrow 1 - \frac{1}{6p} - p + \frac{1}{6} = \frac{1}{3} \Rightarrow \left(1 + \frac{1}{6} - p - \frac{1}{6p}\right) = \frac{1}{3} \Rightarrow \frac{7}{6} - p - \frac{1}{6p} = \frac{1}{3} \] Now: \[ \frac{7}{6} - \frac{1}{3} = p + \frac{1}{6p} \Rightarrow \frac{5}{6} = p + \frac{1}{6p} \] Multiply both sides by \(6p\): \[ 5p = 6p^2 + 1 \Rightarrow 6p^2 - 5p + 1 = 0 \] Now solve: \[ p = \frac{5 \pm \sqrt{25 - 24}}{12} = \frac{5 \pm 1}{12} \Rightarrow p = \frac{6}{12} = \frac{1}{2}, \quad p = \frac{4}{12} = \frac{1}{3} \] Hence, \[ P(A) = \frac{1}{2} \text{ or } \frac{1}{3} \]