Let A and B are two independent events. If the probability that...

P\left(A \cap B\right) = \dfrac{1}{6}$and$P\left(A^{C} \cap B^{C}\right) = \dfrac{1}{3}$</p><p>Now,$P\left(A \cup B\right)C = P\left(A^{C} \cap B^{C}\right) = \dfrac{1}{3}$</p><p>$⇒ 1 - P\left(A \cup B\right) = \dfrac{1}{3}$</p><p>$⇒ P\left(A \cup B\right) = \dfrac{2}{3}$</p><p>But$P \left(A \cup B\right) = P \left(A\right) + $P\left(B\right)$- $P\left(A \cap B\right)$

$⇒ P \left(A\right) + P \left(B\right) = \dfrac{5}{6}... \left(i\right)$

$∵ A$ and $B$ are independent events

$∴ P \left(A \cap B\right) = P \left(A\right)P \left(B\right)$

$⇒ P\left(A\right)P\left(B\right) = \dfrac{1}{6}$

\left[P\left(A\right) -P \left(B\right)\right]2 = \left[P \left(A\right) + P\left(B\right)\right]2 - 4P\left(A\right)P\left(B\right)$</p><p>$= \dfrac{25}{36} - \dfrac{4}{6} = \dfrac{1}{36}$</p><p>$⇒ P \left(A\right) - P \left(B\right) = $\pm$\dfrac{1}{6}$... \left(ii\right) </p><p>On solving Eqs. (i) and (ii), we get</p><p>P\left(A\right) = \dfrac{1}{2}\text{or}\dfrac{1}{3}$

Hence, option (b) is correct.