A box contains 12 identical balls of which 5 are red, 4 blue, and the rest are green. If two balls are selected at random one after the other with replacement, what is the probability that both are red?
\(\frac{25}{144}\)
\(\frac{5}{33}\)
\(\frac{5}{6}\)
\(\frac{103}{132}\)
Correct answer is A
Pr(RR) = \(\frac{5}{12}\) = \(\frac{5}{12} \times \frac{5}{12} = \frac{25}{144}\)