1. Problem: A bag contains 1 red ball, 2 green balls, and 4 yellow balls. A ball is drawn, replaced, and a second ball is drawn. Find probabilities for: (a) both balls are the same colour; (b) no yellow balls are taken out; (c) at least one yellow ball is taken out. 2. Step (a): Calculate probability both balls are the same colour. - Probability both red: $\frac{1}{7} \times \frac{1}{7} = \frac{1}{49}$ - Probability both green: $\frac{2}{7} \times \frac{2}{7} = \frac{4}{49}$ - Probability both yellow: $\frac{4}{7} \times \frac{4}{7} = \frac{16}{49}$ - Total probability same colour: $$\frac{1}{49} + \frac{4}{49} + \frac{16}{49} = \frac{21}{49} = \frac{3}{7}$$ 3. Step (b): Calculate probability no yellow balls are taken out. - Probability first ball not yellow (red or green): $\frac{3}{7}$ - Probability second ball not yellow: $\frac{3}{7}$ - Since replacement, events independent: $$P(\text{no yellow}) = \frac{3}{7} \times \frac{3}{7} = \frac{9}{49}$$ 4. Step (c): Calculate probability at least one yellow ball is taken out. - Use complement rule: $$P(\text{at least one yellow}) = 1 - P(\text{no yellow}) = 1 - \frac{9}{49} = \frac{40}{49}$$ Final answers: - (a) $\frac{3}{7}$ - (b) $\frac{9}{49}$ - (c) $\frac{40}{49}$