Probability

Concept:The random variable $X$ is the absolute difference between the number of Mathematics books and Physics books in a random selection of 6 out of 11 distinct books.The mean is computed by summing each possible value of $X$ multiplied by its probability.Explanation:Total number of ways to choose any 6 books from 11 distinct books: $\binom{11}{6} = 462$.Let $a$ = number of Mathematics books chosen, $b$ = number of Physics books chosen.Since $a + b = 6$, and only 5 Physics books exist, the possible ordered pairs $(a, b)$ are:$(6,0), (5,1), (4,2), (3,3), (2,4), (1,5)$.The corresponding values of $X = |a - b|$ are: $6, 4, 2, 0, 2, 4$.Now compute probabilities for each distinct value of $X$:$X = 6$ occurs only from $(6,0)$:$P(X=6) = \dfrac{\binom{6}{6} \cdot \binom{5}{0}}{\binom{11}{6}} = \dfrac{1}{462}$.$X = 4$ occurs from $(5,1)$ and $(1,5)$:$P(X=4) = \dfrac{\binom{6}{5} \cdot \binom{5}{1} + \binom{6}{1} \cdot \binom{5}{5}}{\binom{11}{6}} = \dfrac{6\cdot5 + 6\cdot1}{462} = \dfrac{36}{462}$.$X = 2$ occurs from $(4,2)$ and $(2,4)$:$P(X=2) = \dfrac{\binom{6}{4} \cdot \binom{5}{2} + \binom{6}{2} \cdot \binom{5}{4}}{\binom{11}{6}} = \dfrac{15\cdot10 + 15\cdot5}{462} = \dfrac{225}{462}$.$X = 0$ occurs from $(3,3)$:$P(X=0) = \dfrac{\binom{6}{3} \cdot \binom{5}{3}}{\binom{11}{6}} = \dfrac{20\cdot10}{462} = \dfrac{200}{462}$.Mean $\alpha$ of $X$ is:$$\alpha = \sum X \cdot P(X) = \frac{6 \cdot 1 + 4 \cdot 36 + 2 \cdot 225 + 0 \cdot 200}{462}$$Simplify the numerator: $6 + 144 + 450 = 600$.Thus $\alpha = \dfrac{600}{462} = \dfrac{100}{77}$.Therefore, $77\alpha = 77 \times \dfrac{100}{77} = 100$.Answer:$100$