Answer to Question #326180 in Statistics and Probability for Robera

Question

Two dice are rolled. Let X be a random variable denoting the sum of the numbers on

the two dice.

i) Give the probability distribution of X

ii) Compute the expected value of X and its variance

Answer 1

X={2,3,4,5,6,7,8,9,10,11,12}

Let (a,b) be a set where a - number on the first die, b - on the second die, then

X=2: (1,1) $P(X=2)={\frac 1 {36}}$

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

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

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

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

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

X=8; (2,6) (3,5) (4,4) (5,3) (6,2) $P(X=7)={\frac 5 {36}}$

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

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

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

X=12; (6,6) $P(X=12)={\frac 1 {36}}$

Distribution is symmetrical about X=7, so

M(X)=7

$D(X)=M(X^2)-M^2(X)={\frac {329} 6}-49={\frac {35} 6}$