1. **State the problem:** A bag contains 8 identical red balls and an unknown number of white balls. The probability of picking a red ball initially is given as $\frac{5}{2}$, which is impossible since probabilities must be between 0 and 1. We will assume the problem meant the probability is $\frac{2}{5}$ instead. 2. **Define variables:** Let the number of white balls be $w$. 3. **Initial probability formula:** The probability of picking a red ball is $$P(\text{red}) = \frac{\text{number of red balls}}{\text{total number of balls}} = \frac{8}{8 + w} = \frac{2}{5}.$$ 4. **Solve for $w$:** $$\frac{8}{8 + w} = \frac{2}{5}$$ Cross multiply: $$8 \times 5 = 2 \times (8 + w)$$ $$40 = 16 + 2w$$ Subtract 16 from both sides: $$40 - 16 = 2w$$ $$24 = 2w$$ Divide both sides by 2: $$w = 12.$$ 5. **After adding 4 white balls:** New number of white balls = $12 + 4 = 16$. Total balls now = $8 + 16 = 24$. 6. **New probability of picking a red ball:** $$P(\text{red new}) = \frac{8}{24} = \frac{1}{3}.$$ **Final answer:** The probability of picking a red ball after adding 4 white balls is $\frac{1}{3}$.