Two dice are rolled simultaneously. Find the probability distribution…

The correct answer is Option (2) → 

$X$	2	3	4	5	6	7	8	9	10	11	12
$P(X)$	$\frac{1}{36}$	$\frac{1}{18}$	$\frac{1}{12}$	$\frac{1}{9}$	$\frac{5}{36}$	$\frac{1}{6}$	$\frac{5}{36}$	$\frac{1}{9}$	$\frac{1}{12}$	$\frac{1}{18}$	$\frac{1}{36}$

The sample space associated with this experiment is

$S = \{(1, 1), (1, 2), (1, 3), ..., (1, 6), (2, 1), (2, 2) ..., (2, 6), (3, 1), ..., (3, 6), (4, 1), ..., (4, 6), (5, 1), ..., (5, 6), (6, 1), ..., (6, 6)\}$.

All 36 outcomes are equally likely.

Let random variable X be defined as the sum of digits on the two dice, then X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

$P(X = 2) = P\{(1, 1)\}=\frac{1}{36}$

$P(X = 3) = P\{(1, 2), (2, 1)\}=\frac{2}{36}=\frac{1}{18}$

$P(X = 4) = P\{(1, 3), (3, 1), (2, 2)\} =\frac{3}{36}=\frac{1}{12}$

$P(X = 5) = P\{(1, 4), (4, 1), (2, 3), (3, 2)\}=\frac{4}{36}=\frac{1}{9}$

$P(X=6)= P\{(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)\} =\frac{5}{36}$

$P(X = 7) = P\{(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)\}=\frac{6}{36}=\frac{1}{6}$

$P(X = 8) = P\{(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)\} =\frac{5}{36}$

$P(X = 9)= P\{(3, 6), (6, 3), (4, 5), (5, 4)\} =\frac{4}{36}=\frac{1}{9}$

$P(X = 10) = P\{(4, 6), (6, 4), (5, 5)\} =\frac{3}{36}=\frac{1}{12}$

$P(X = 11) = P\{(5, 6), (6,5)\}=\frac{2}{36}=\frac{1}{18}$

$P(X = 12) = P\{(6, 6)\} =\frac{1}{36}$

Hence, the required probability distribution is

$X$	2	3	4	5	6	7	8	9	10	11	12
$P(X)$	$\frac{1}{36}$	$\frac{1}{18}$	$\frac{1}{12}$	$\frac{1}{9}$	$\frac{5}{36}$	$\frac{1}{6}$	$\frac{5}{36}$	$\frac{1}{9}$	$\frac{1}{12}$	$\frac{1}{18}$	$\frac{1}{36}$