1. **State the problem:** We have two boxes. The first box contains 4 red and 5 white balls, and the second box contains 6 red and 3 white balls. We pick one ball randomly from each box. We want to find the probability that at least one of the balls picked is red. 2. **Formula and approach:** The probability of "at least one" event happening is easier to find by using the complement rule: $$P(\text{at least one red}) = 1 - P(\text{no red balls})$$ 3. **Calculate the probability of no red balls:** - Probability of picking a white ball from the first box: $$\frac{5}{9}$$ (since total balls = 4 + 5 = 9) - Probability of picking a white ball from the second box: $$\frac{3}{9} = \frac{1}{3}$$ (since total balls = 6 + 3 = 9) 4. **Calculate the combined probability of no red balls:** Since the picks are independent, multiply the probabilities: $$P(\text{no red balls}) = \frac{5}{9} \times \frac{1}{3} = \frac{5}{27}$$ 5. **Calculate the probability of at least one red ball:** $$P(\text{at least one red}) = 1 - \frac{5}{27} = \frac{22}{27}$$ **Final answer:** The probability that at least one ball is red is $$\frac{22}{27}$$.