Question 11

Total ways to choose 6 books from 11 is $\binom{11}{6} = 462$. Let $M$ and $P$ be the number of Math and Physics books chosen respectively, where $M + P = 6$. The possible scenarios are:

1. $M=6, P=0 \Rightarrow X=|6-0|=6$. Ways: $\binom{6}{6}\binom{5}{0} = 1$.
2. $M=5, P=1 \Rightarrow X=|5-1|=4$. Ways: $\binom{6}{5}\binom{5}{1} = 30$.
3. $M=4, P=2 \Rightarrow X=|4-2|=2$. Ways: $\binom{6}{4}\binom{5}{2} = 150$.
4. $M=3, P=3 \Rightarrow X=|3-3|=0$. Ways: $\binom{6}{3}\binom{5}{3} = 200$.
5. $M=2, P=4 \Rightarrow X=|2-4|=2$. Ways: $\binom{6}{2}\binom{5}{4} = 75$.
6. $M=1, P=5 \Rightarrow X=|1-5|=4$. Ways: $\binom{6}{1}\binom{5}{5} = 6$. Note: $M=0, P=6$ is impossible as only 5 Physics books exist.

Mean $\alpha = E[X] = \frac{1}{462} \sum X_i \cdot \text{ways}_i$ $\alpha = \frac{1}{462} [6(1) + 4(30) + 2(150) + 0(200) + 2(75) + 4(6)]$ $\alpha = \frac{1}{462} [6 + 120 + 300 + 0 + 150 + 24] = \frac{600}{462}$ $77\alpha = 77 \times \frac{600}{462} = \frac{600}{6} = 100$. Final Answer: 100.