Question

Part 4: Applications ie is modified so that the 'l'is now a '4'. This modified dice is thrown twice and the sum of the two throws recorded. Give a probability distribution table and also the expected value

Answer 1

let X1 and X2 are outcome on two throws:

below is pmf of total X:

P(X=4)=P(2,2)=(1/6)+(1/6)=1/36 ( as there is one outcome (2,2) which result into sum as 4)

P(X=5)=P(2,3)+P(3,2)=(1/6)*(1/6)+(1/6)*(1/6)=2/36

P(X=6)=P(2,4)+P(3,3)+P(4,2)=(1/6)*(2/6)+(1/6)*(1/6)+(2/6)*(1/6)=5/36

P(X=7)=P(2,5)+P(3,4)+P(4,3)+(5,2)=(1/36)+(2/36)+(2/36)+(1/36)=6/36

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+(6.2)=1/36+1/36+4/36+1/36+1/36=8/36

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=(1/36)+2/36+2/36+1/36=6/36

P(X=10)=P(4,6)+P(5,5)+P(6,4)=2/36+1/36+2/36=5/36

P(X=11)=P(5,6)+P(6,5)=(1/36+1/36)=2/36

P(X=12)=P(6,6)=1/36

\\expected value =E(X)=$\sum$xP(x)=4*(1/36)+5*(2/36)+6*(5/36)+7*(6/36)+8*(8/36)+9*(6/36)+10*(5/36)+11*(2/36)+12*(1/36)

=288/36=8