Question

A bag contains 3 red, 4 white and 5 blue balls. If two balls are drawn at random, then the probability that they are of different colours is:

• A. $\frac{47}{66}$
• B. $\frac{23}{33}$
• C. $\frac{47}{132}$
• D. $\frac{47}{33}$
• E. $\frac{70}{33}$

Hint:

When asked for "different colors" with more than two types, always consider the "1 minus same color" approach. It involves 3 simple combinations instead of calculating Red-White, Red-Blue, and White-Blue pairs separately.

Answer 1

The Correct Option is A

Concept: The probability of an event is the number of favorable outcomes divided by the total number of outcomes. When drawing multiple items, it is often easier to use the complement: $P(\text{different colors}) = 1 - P(\text{same color})$.

Step 1: Calculate the total number of ways to draw 2 balls.
Total balls $= 3 + 4 + 5 = 12$.
Total outcomes $= \binom{12}{2} = \frac{12 \times 11}{2} = 66$.

Step 2: Calculate the number of ways to draw 2 balls of the same color.

• Both Red: $\binom{3}{2} = 3$
• Both White: $\binom{4}{2} = 6$
• Both Blue: $\binom{5}{2} = 10$ Total same color outcomes $= 3 + 6 + 10 = 19$.

Step 3: Find the probability of different colors.
Number of favorable outcomes (different colors) $= 66 - 19 = 47$. \[ P(\text{different}) = \frac{47}{66} \]