1. **Stating the problem:** We want to find the probability that two customers selected at random both require a takeout box. 2. **Understanding the data:** Total customers = 27 + 28 + 33 + 12 = 100. Customers who require a takeout box = 27 + 28 = 55. 3. **Formula for probability of both requiring a takeout box:** When selecting two customers without replacement, the probability both require a takeout box is: $$P = \frac{55}{100} \times \frac{54}{99}$$ 4. **Calculate the probability:** $$P = \frac{55 \times 54}{100 \times 99} = \frac{2970}{9900}$$ 5. **Simplify the fraction:** Divide numerator and denominator by 30: $$\frac{\cancel{2970}^{99}}{\cancel{9900}^{330}}$$ Divide numerator and denominator by 3: $$\frac{\cancel{99}^{33}}{\cancel{330}^{110}}$$ Divide numerator and denominator by 11: $$\frac{\cancel{33}^{3}}{\cancel{110}^{10}}$$ So the simplified probability is: $$\frac{3}{10}$$ 6. **Check the answer:** The user mentioned the answer is $\frac{59}{169}$, which is approximately 0.3497. Our calculation $\frac{3}{10} = 0.3$ is different, so let's re-express the problem carefully. 7. **Re-examining the problem:** Total customers = 100. Number requiring takeout box = 27 + 28 = 55. Probability first customer requires takeout box = $\frac{55}{100}$. After selecting one, remaining customers = 99. Remaining requiring takeout box = 54. Probability second customer requires takeout box = $\frac{54}{99}$. Therefore: $$P = \frac{55}{100} \times \frac{54}{99} = \frac{2970}{9900} = \frac{33}{110} = \frac{3}{10}$$ This matches our previous result. 8. **Alternative approach:** If the user’s answer is $\frac{59}{169}$, it might be based on a different total or selection method. 9. **Conclusion:** The probability that both customers require a takeout box is: $$\boxed{\frac{3}{10}}$$