1. **Problem Statement:** We have three girls Aisha, Betty, and Cate packing juice bottles. The probabilities that a crate is packed by Aisha, Betty, and Cate are 55%, 30%, and 15% respectively. The probabilities that a crate packed by Aisha, Betty, and Cate is broken are 0.7, 0.2, and 0.1 respectively. We want to find the probability that a crate with broken bottles was packed by Aisha using Bayes' theorem. 2. **Bayes' Theorem Formula:** $$P(A|B) = \frac{P(B|A) \times P(A)}{P(B)}$$ where: - $P(A|B)$ is the probability that Aisha packed the crate given it is broken. - $P(B|A)$ is the probability the crate is broken given Aisha packed it. - $P(A)$ is the probability that Aisha packed the crate. - $P(B)$ is the total probability that a crate is broken. 3. **Given values:** $$P(A) = 0.55, \quad P(B|A) = 0.7$$ $$P(Betty) = 0.30, \quad P(B|Betty) = 0.2$$ $$P(Cate) = 0.15, \quad P(B|Cate) = 0.1$$ 4. **Calculate total probability of broken crate $P(B)$ using law of total probability:** $$P(B) = P(B|A)P(A) + P(B|Betty)P(Betty) + P(B|Cate)P(Cate)$$ $$= 0.7 \times 0.55 + 0.2 \times 0.30 + 0.1 \times 0.15$$ $$= 0.385 + 0.06 + 0.015 = 0.46$$ 5. **Apply Bayes' theorem:** $$P(A|B) = \frac{0.7 \times 0.55}{0.46} = \frac{0.385}{0.46}$$ 6. **Simplify fraction:** $$P(A|B) = \frac{\cancel{0.385}}{\cancel{0.46}} = 0.837$$ **Final answer:** The probability that a crate with broken bottles was packed by Aisha is approximately **0.837** or **83.7%**.