Question

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

• A. 31
• B. 36
• C. 48
• D. 72

Answer 1

The Correct Option is C

\( \alpha - \beta \)	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:

\( (5^0 + 5^1)(7^0 + 7^1) = 6 \times 8 = 48 \)