Two fair dices, each numbered 1, 2, ..., 6, are tossed together. Find the probability that they both show even numbers.
\(\frac{1}{3}\)
\(\frac{1}{4}\)
\(\frac{1}{6}\)
\(\frac{1}{12}\)
Correct answer is B
P(even in 1 dice) = \(\frac{3}{6} = \frac{1}{2}\)
P(even in 2 fair die) = \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)