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 the two balls are red?
-
\(\frac{5}{33}\)
- \(\frac{5}{36}\)
- \(\frac{103}{132}\)
- \(\frac{31}{36}\)
The correct answer is: A
Explanation
Total number of balls = 5 + 3 + 4
= 12 balls
P(first ball is red) = \(\frac{5}{12}\)
P(second ball is red) = \(\frac{4}{11}\)
\(\therefore\) P(both balls are red) = \(\frac{5}{12} \times \frac{4}{11}\)
= \(\frac{5}{33}\)