A box contains 5 red, 3 green and 4 blue balls. A boy is allowed to take away two balls from the box. What is the probability that one is green and the other is blue?
\(\frac{2}{11}\)
\(\frac{5}{12}\)
\(\frac{8}{12}\)
\(\frac{7}{11}\)
Correct answer is A
Total number of balls = 5 + 3 + 4 = 12 balls
P(one ball is green and the other is blue) = P(first ball is green and second blue) + P(first ball is blue and the second green)
= \(\frac{3}{12} \times \frac{4}{11} + \frac{4}{12} \times \frac{3}{11}\)
= \(\frac{2}{11}\)