▶️ Answer/Explanation
Solution
(a) Calculating \( P(A \cap B) \):
\[ P(A) = \frac{4}{6} = \frac{2}{3} \]
After drawing one red ball, there are 5 balls left (3 red and 2 green):
\[ P(B|A) = \frac{2}{5} \]
\[ P(A \cap B) = P(A) \times P(B|A) = \frac{2}{3} \times \frac{2}{5} = \frac{4}{15} \]
✅ Answer: \(\frac{4}{15}\)
(b) Calculating \( P(C) \):
There are two ways to get exactly one red ball:
1. First ball red and second ball green: \( \frac{4}{6} \times \frac{2}{5} = \frac{4}{15} \)
2. First ball green and second ball red: \( \frac{2}{6} \times \frac{4}{5} = \frac{4}{15} \)
Total probability: \( \frac{4}{15} + \frac{4}{15} = \frac{8}{15} \)
✅ Answer: \(\frac{8}{15}\)