Question

Consider that you toss a fair 6-sided die containing the numbers 1-2-3-4-5-6 and also toss a fair 4-sided die containing the numbers 1-2-3-4. Find the probability distribution for the sum of the values on the two dice. Also, find the mean and the variance of this probability distribution.

Please provide a well written and well explained answer.

Answer 1

let

X is number on 6 sided die

So P(X=1)=P(X=2)=P(X=3)=P(X=4)=P(X=5)=P(X=6) =1/6

Y is number on 4 sided die

so

P(Y=1)=P(Y=2)=P(Y=3)=P(Y=4)

we have to find the Distribution of T=X+Y

P(T=2) =P(X=1,Y=1) =P(X=1)P(Y=1) =(1/6)(1/4) =1/24

P(T=3) =P(X=1,Y=2) +P(X=2,Y=1) =P(X=1)P(Y=2) +P(X=2)P(Y=1) =(1/6)*(1/4) +(1/6)*(1/4) =2/24

P(T=4) =P(X=1,Y=3)+P(X=3,Y=1) +P(X=2,Y=2) =(1/6)(1/4)+(1/6)(1/4)+(1/6)(1/4)=3/24

P(T=5) =P(X=1,Y=4)+P(X=4,Y=1)+P(X=2,Y=3)+P(X=3,Y=2)=(1/6)(1/4)+(1/6)(1/4)+(1/6)(1/4)+(1/6)(1/4)=4/24

P(T=6) =P(X=5,Y=1)+P(X=2,Y=4)+P(X=4,Y=2)+P(X=3,Y=3)=(1/6)(1/4)+(1/6)(1/4)+(1/6)(1/4)+(1/6)(1/4)=4/24

P(T=7) =P(X=6,Y=1)+P(X=5,Y=2)+P(X=4,Y=3)+P(X=3,Y=4) =(1/6)(1/4)+(1/6)(1/4)+(1/6)(1/4)+(1/6)(1/4)=4/24

P(T=8) =P(X=6,Y=2)+P(X=5,Y=3)+P(X=4,Y=4)=(1/6)(1/4)+(1/6)(1/4)+(1/6)(1/4)=3/24

P(T=9)=P(X=6,Y=3)+P(X=5,Y=4) =(1/6)(1/4)+(1/6)(1/4)=2/24

P(T=10) =P(X=6,Y=4) =(1/6)*(1/4) =1/24

Now

Mean =E(T) =Sum of (t*P(T=t))

=2*(1/24)+3*(2/24)+4*(3/24) +5*(4/24)+6*(4/24)+7*(4/24)+8*(3/24)+9*(2/24)+10*(1/24)

=(1/24) ( 2+6+12+20+24+28+24+18+10) =144/24 =6

Now

E(T²)=sum of (t²*P(T=t))

=4*(1/24)+9*(2/24)+16*(3/24) +25*(4/24)+36*(4/24)+49*(4/24)+64*(3/24)+81*(2/24)+100*(1/24)

=(1/24) (4+18+48+100+144+196+192+162+100) =964/24 =40.167

Now

Variance(T) =E(T²)-(E(T))²

=40.167-36 =4.167