Question

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 {

Hint:

Let k be the number of Mathematics books chosen out of the 6 selected, so the number of Physics books chosen is automatically 6 minus k, and the random variable X becomes a simple expression written in terms of k alone. Set up the probability of each value of k using combinations, choosing k books from the 6 Mathematics books and the rest from the 5 Physics books, out of all the ways to choose 6 books from 11 in total. Work out X for every possible value of k first, since the same value of X can come from more than one value of k, before you bring in the probabilities to find the mean.

Answer 1

Correct Answer: 100

Step 1: Understanding the Question:
This is a probability distribution problem. We need to find the expected value (mean) of a random variable $X$ defined by the composition of the sample. Let $k$ be the number of Mathematics books selected. Then the number of Physics books is $6-k$. $X = |k - (6-k)| = |2k-6|$.
Step 2: Key Formula or Approach:

Mean $\alpha = E[X] = \sum x_i \cdot P(X = x_i)$.

$P(k) = \frac{\binom{6}{k} \binom{5}{6-k}}{\binom{11}{6}}$.

Step 3: Detailed Explanation:

Total ways to choose 6 books from 11: $\binom{11}{6} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$.

Possible values of $k$ (Math books) are $1, 2, 3, 4, 5, 6$ (since total Physics books is only 5).

For $k=1$, $X = |2-6|=4$. Prob $P(1) = \frac{\binom{6}{1}\binom{5}{5}}{462} = \frac{6 \times 1}{462} = \frac{6}{462}$.

For $k=2$, $X = |4-6|=2$. Prob $P(2) = \frac{\binom{6}{2}\binom{5}{4}}{462} = \frac{15 \times 5}{462} = \frac{75}{462}$.

For $k=3$, $X = |6-6|=0$. Prob $P(3) = \frac{\binom{6}{3}\binom{5}{3}}{462} = \frac{20 \times 10}{462} = \frac{200}{462}$.

For $k=4$, $X = |8-6|=2$. Prob $P(4) = \frac{\binom{6}{4}\binom{5}{2}}{462} = \frac{15 \times 10}{462} = \frac{150}{462}$.

For $k=5$, $X = |10-6|=4$. Prob $P(5) = \frac{\binom{6}{5}\binom{5}{1}}{462} = \frac{6 \times 5}{462} = \frac{30}{462}$.

For $k=6$, $X = |12-6|=6$. Prob $P(6) = \frac{\binom{6}{6}\binom{5}{0}}{462} = \frac{1 \times 1}{462} = \frac{1}{462}$.

$\alpha = \frac{1}{462} [4(6) + 2(75) + 0(200) + 2(150) + 4(30) + 6(1)]$.
$\alpha = \frac{24 + 150 + 0 + 300 + 120 + 6}{462} = \frac{600}{462}$.

Simplify $\alpha$: $\alpha = \frac{100}{77}$.

Therefore, $77\alpha = 77 \times \frac{100}{77} = 100$.

Step 4: Final Answer:
The value of $77\alpha$ is 100.

Answer 2

Concept:

• Organise the six possible cases into a clear table with one row per case, so the pattern of the calculation is visible at a glance instead of buried in separate probability statements.

Step 1: Note the total ways to pick 6 books out of 11.
$\binom{11}{6} = \dfrac{11!}{6!\,5!} = 462$

Step 2: Tabulate every case.

Maths (k)	Physics (6-k)	X=|2k-6|	Ways = C(6,k)·C(5,6-k)
1	5	4	6
2	4	2	75
3	3	0	200
4	2	2	150
5	1	4	30
6	0	6	1

Step 3: Multiply each $X$ by its Ways column and add.
$\sum X\cdot\text{Ways} = 1(4)(6)+2(75)+0(200)+2(150)+4(30)+6(1) = 24+150+0+300+120+6=600$

Step 4: Divide by the total to get the mean.
$\alpha = \dfrac{600}{462} = \dfrac{100}{77}$

Step 5: Multiply by 77.
$77\alpha = 100$

Final Answer: $77\alpha = 100$