45. A bag contains 3 blue, 2 green and 5 red balls. If four balls are picked at random, what is the probability that two are green and two are blue?

### Concept & Probability by Combinations To find the probability of multiple events, we use combinations to find the number of favorable outcomes and divide by the total possible outcomes. $$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$$ ### Step-by-Step Solution * **Total number of balls:** $3 \text{ (blue)} + 2 \text{ (green)} + 5 \text{ (red)} = 10 \text{ balls}$. * **Total possible outcomes:** Selecting 4 balls out of 10. $n(S) = ^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$. * **Favorable outcomes:** We need exactly 2 green and 2 blue balls. Ways to select 2 green balls from 2: $^2C_2 = 1$. Ways to select 2 blue balls from 3: $^3C_2 = \frac{3 \times 2}{2 \times 1} = 3$. * Total favorable outcomes $n(E) = 1 \times 3 = 3$. * **Probability:** $P(E) = \frac{n(E)}{n(S)} = \frac{3}{210} = \frac{1}{70}$. ### Exam Strategy & Shortcut Whenever the number of items required equals the total number of that item (like 2 green out of 2 green), that part of the combination is always 1. This saves calculation time. So, focus on calculating $^3C_2$ and $^{10}C_4$. ### Common Pitfall A common mistake is adding the combinations ($^2C_2 + ^3C_2$) instead of multiplying them. Since we need both green AND blue balls simultaneously, the fundamental principle of counting dictates we must multiply. ### Final Answer Therefore, the correct answer is **$\frac{1}{70}$**.