Two dice A and B are rolled, Let the numbers obtained on A and B be α and β respectively. If the variance of α - β is \(\frac{p}{q}\), where p and q are coprime, then the sum of the positive divisors of p is equal to
- 31
- 36
- 48
- 72
The Correct Option is C
Solution and Explanation
| \( \alpha - \beta \) | Case | \( P \) |
|---|---|---|
| 5 | (6, 1) | \(\frac{1}{36}\) |
| 4 | (6, 2), (5, 1) | \(\frac{2}{36}\) |
| 3 | (6, 3), (5, 2), (4, 1) | \(\frac{3}{36}\) |
| 2 | (6, 4), (5, 3), (4, 3), (3, 1) | \(\frac{4}{36}\) |
| 1 | (6, 5), (5, 4), (4, 3), (3, 2), (2, 1) | \(\frac{5}{36}\) |
| 0 | (6, 6), (5, 5), ..., (1, 1) | \(\frac{6}{36}\) |
| -1 | --- | \(\frac{5}{36}\) |
| -2 | --- | \(\frac{4}{36}\) |
| -3 | --- | \(\frac{3}{36}\) |
| -4 | (2, 6), (1, 5) | \(\frac{2}{36}\) |
| -5 | (1, 6) | \(\frac{1}{36}\) |
Step 1: Calculation of \( \Sigma(x^2) \)
\( \Sigma(x^2) = \Sigma x^2 P(x) = 2 \left( \frac{25}{36} + \frac{32}{36} + \frac{27}{36} + \frac{16}{36} + \frac{5}{36} \right) \)
Simplify:
\( \Sigma(x^2) = \frac{105}{18} = \frac{35}{6} \)
Step 2: Mean (\( \mu \))
As the data is symmetric:
\( \mu = \Sigma(x) = 0 \)
Step 3: Variance (\( \sigma^2 \))
\( \sigma^2 = \Sigma(x^2 P(x)) = \frac{35}{6} \)
The total probability \( P \) is:
\( P = 35 = 5 \times 7 \)
Step 4: Sum of divisors of \( P \)
The sum of divisors of 35 is:
\( (5^0 + 5^1)(7^0 + 7^1) = 6 \times 8 = 48 \)
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