Question

From a set \(\{1,2,3,4,5,6,7\}\) two numbers are selected at random. The random variable \(X\) is defined as the absolute difference of the numbers selected. The mean of \(X\) is

• A. \(21\)
• B. \(4\)
• C. \(\frac{8}{3}\)
• D. \(\frac{7}{2}\)

Hint:

For a set of consecutive integers from \(1\) to \(n\), the expected absolute difference of two randomly chosen numbers is always given by the shortcut formula \(\frac{n+1}{3}\).
Here, \(\frac{7+1}{3} = \frac{8}{3}\).

Answer 1

The Correct Option is C

Concept:
• Total number of ways to select 2 numbers from \(n\) elements is \(\binom{n}{2}\).
• The absolute difference \(X\) can take integer values from \(1\) to \(n-1\).
• The number of pairs with a absolute difference of \(d\) from a set of \(n\) consecutive integers is exactly \(n - d\).
• The mean (expectation) of \(X\) is given by \(\text{E}[X] = \sum d \cdot \text{P}(X=d)\).

Step 1: Calculate total outcomes and possible values of \(X\)
Total pairs: \(\binom{7}{2} = \frac{7 \times 6}{2} = 21\).
Possible values for difference \(d\): \(\{1, 2, 3, 4, 5, 6\}\).

Step 2: Find the frequency and probability distribution for each difference \(d\)
Number of pairs for difference \(d\) is \(7 - d\):
• \(\text{P}(X=1) = \frac{7-1}{21} = \frac{6}{21}\)
• \(\text{P}(X=2) = \frac{7-2}{21} = \frac{5}{21}\)
• \(\text{P}(X=3) = \frac{7-3}{21} = \frac{4}{21}\)
• \(\text{P}(X=4) = \frac{7-4}{21} = \frac{3}{21}\)
• \(\text{P}(X=5) = \frac{7-5}{21} = \frac{2}{21}\)
• \(\text{P}(X=6) = \frac{7-6}{21} = \frac{1}{21}\)

Step 3: Calculate the mean \(\text{E}[X]\)
\[ \text{E}[X] = \frac{1}{21} \sum_{d=1}^{6} d \cdot (7 - d) \] \[ = \frac{1}{21} [1(6) + 2(5) + 3(4) + 4(3) + 5(2) + 6(1)] \] \[ = \frac{1}{21} [6 + 10 + 12 + 12 + 10 + 6] = \frac{56}{21} = \frac{8}{3} \]