A bookshelf contains 6 distinct books of Mathematics and 5 distinct books of Physics. From these 11 books, 6 books are chosen at random.

Let $X$ be the absolute value of the difference between the number of Mathematics books chosen and the number of Physics books chosen.

If $\alpha$ is the mean of the random variable $X$, then the value of $77\alpha$ is .

The correct answer is 100.

Step 1: Define the sample space and variables
We are given a total of 11 distinct books (6 Mathematics books and 5 Physics books), out of which 6 books are chosen at random.

Let $k$ be the number of Mathematics books chosen from the 6 Mathematics books. Then, the number of Physics books chosen from the 5 Physics books is $6-k$.

Since there are only 5 Physics books available in total, the maximum number of Physics books that can be chosen is 5. Therefore:

$6-k\le 5\Rightarrow k\ge 1$

Thus, the possible number of Mathematics books chosen, $k$, can be 1, 2, 3, 4, 5, or 6.

The total number of ways to choose 6 books from 11 is given by the combination formula:

$\text{Total ways}=\left(\begin{array}{}116\end{array}\right)=\frac{11\times 10\times 9\times 8\times 7}{5\times 4\times 3\times 2\times 1}=462$

Step 2: Determine the values of random variable $X$
The random variable $X$ represents the absolute difference between the number of Mathematics books and Physics books chosen:

$X=|k-(6-k\left)\right|=|2k-6|$

We evaluate $X$ and its corresponding probability $P\left(X\right)$ for each possible value of $k$:

1. For $k=1$ (1 Math, 5 Physics):
$X=\left|2\right(1)-6|=4$
$P(k=1)=\frac{\left(\begin{array}{}61\end{array}\right)\left(\begin{array}{}55\end{array}\right)}{462}=\frac{6\times 1}{462}=\frac{6}{462}$

2. For $k=2$ (2 Math, 4 Physics):
$X=\left|2\right(2)-6|=2$
$P(k=2)=\frac{\left(\begin{array}{}62\end{array}\right)\left(\begin{array}{}54\end{array}\right)}{462}=\frac{15\times 5}{462}=\frac{75}{462}$

3. For $k=3$ (3 Math, 3 Physics):
$X=\left|2\right(3)-6|=0$
$P(k=3)=\frac{\left(\begin{array}{}63\end{array}\right)\left(\begin{array}{}53\end{array}\right)}{462}=\frac{20\times 10}{462}=\frac{200}{462}$

4. For $k=4$ (4 Math, 2 Physics):
$X=\left|2\right(4)-6|=2$
$P(k=4)=\frac{\left(\begin{array}{}64\end{array}\right)\left(\begin{array}{}52\end{array}\right)}{462}=\frac{15\times 10}{462}=\frac{150}{462}$

5. For $k=5$ (5 Math, 1 Physics):
$X=\left|2\right(5)-6|=4$
$P(k=5)=\frac{\left(\begin{array}{}65\end{array}\right)\left(\begin{array}{}51\end{array}\right)}{462}=\frac{6\times 5}{462}=\frac{30}{462}$

6. For $k=6$ (6 Math, 0 Physics):
$X=\left|2\right(6)-6|=6$
$P(k=6)=\frac{\left(\begin{array}{}66\end{array}\right)\left(\begin{array}{}50\end{array}\right)}{462}=\frac{1\times 1}{462}=\frac{1}{462}$

Step 3: Calculate the mean $\alpha =E\left[X\right]$
The expected value (mean) of $X$ is calculated as:

$\alpha =\sum X·P\left(X\right)$

$\alpha =\left(4\times \frac{6}{462}\right)+\left(2\times \frac{75}{462}\right)+\left(0\times \frac{200}{462}\right)+\left(2\times \frac{150}{462}\right)+\left(4\times \frac{30}{462}\right)+\left(6\times \frac{1}{462}\right)$

$\alpha =\frac{24+150+0+300+120+6}{462}$

$\alpha =\frac{600}{462}=\frac{100}{77}$

Step 4: Find the value of $77\alpha$
Finally, multiplying by 77 gives:

$77\alpha =77\times \frac{100}{77}=100$