The probability of an event P is \(\frac{3}{4}\) while that of another event Q is \(\frac{1}{6}\). If the probability of both P and Q is \(\frac{1}{2}\). What is the probability of either P or Q.

A.

\(\frac{1}{96}\)

B.

\(\frac{1}{8}\)

C.

\(\frac{5}{6}\)

D.

\(\frac{11}{12}\)

Correct answer is D

Prob of P = \(\frac{3}{4}\)

Prob of Q = \(\frac{1}{6}\)

Prob of both P or Q = \(\frac{3}{4} + \frac{1}{6} = \frac{11}{12}\)