Question

There are two octahedral dice with faces marked 1 through 8. They are constructed such that the side 8 is 1.5 times as probable as the 2 through 7 and the sum of the probabilities of the 1 and the 8 equals that of the pairs on opposing faces. Let x be the sum of the faces when we roll the two dice.

Find the following:

a. Probability distribution of x

b. Mean of x

c. Variance of x

Answer 1

Let the probability of $1$ be $y.$ Let the probability of $2$ be $s.$ Then

$y+s+s+s+s+s+s+1.5s=1$ $y+1.5s=s+s$

$y+7.5s=1$ $y=0.5s$

$y=1/16, s=1/8$

There are $8^2=64$ possible outcomes for $X$

$1+1=2$

$1+2=2+1=3$

$1+3=2+2=3+1=4$

$1+4=2+3=3+2=4+1=5$

$1+5=2+4=3+3=4+2=5+1=6$

$1+6=2+5=3+4=4+3=5+2=6+1=7$

$1+7=2+6=3+5=4+4=5+3=6+2$ $=7+1=8$

$1+8=2+7=3+6=4+5=5+4=6+3$ $=7+2=8+1=9$

$2+8=3+7=4+6=5+5=6+4=7+3$ $=8+2=10$

$3+8=4+7=5+6=6+5=7+4$ $=8+3=11$

$4+8=5+7=6+6=7+5=8+4=12$

$5+8=6+7=7+6=8+5=13$

$6+8=7+7=8+6=14$

$7+8=8+7=15$

$8+8=16$

$P(X=2)=\dfrac{1}{16}(\dfrac{1}{16})=\dfrac{1}{256}$

$P(X=3)=\dfrac{1}{16}(\dfrac{1}{8})+\dfrac{1}{8}(\dfrac{1}{16})=\dfrac{1}{64}$

$P(X=4)=\dfrac{1}{16}(\dfrac{1}{8})+\dfrac{1}{8}(\dfrac{1}{8})+\dfrac{1}{8}(\dfrac{1}{16})=\dfrac{1}{32}$

$P(X=5)=\dfrac{1}{16}(\dfrac{1}{8})+2(\dfrac{1}{8})(\dfrac{1}{8})+\dfrac{1}{8}(\dfrac{1}{16})=\dfrac{3}{64}$

$P(X=6)=\dfrac{1}{16}(\dfrac{1}{8})+3(\dfrac{1}{8})(\dfrac{1}{8})+\dfrac{1}{8}(\dfrac{1}{16})=\dfrac{1}{16}$

$P(X=7)=\dfrac{1}{16}(\dfrac{1}{8})+4(\dfrac{1}{8})(\dfrac{1}{8})+\dfrac{1}{8}(\dfrac{1}{16})=\dfrac{5}{64}$

$P(X=8)=\dfrac{1}{16}(\dfrac{1}{8})+5(\dfrac{1}{8})(\dfrac{1}{8})+\dfrac{1}{8}(\dfrac{1}{16})=\dfrac{3}{32}$

$P(X=9)=\dfrac{1}{16}(\dfrac{3}{16})+6(\dfrac{1}{8})(\dfrac{1}{8})+\dfrac{3}{16}(\dfrac{1}{16})=\dfrac{15}{128}$

$P(X=10)=\dfrac{1}{8}(\dfrac{3}{16})+5(\dfrac{1}{8})(\dfrac{1}{8})+\dfrac{3}{16}(\dfrac{1}{8})=\dfrac{1}{8}$

$P(X=11)=\dfrac{1}{8}(\dfrac{3}{16})+4(\dfrac{1}{8})(\dfrac{1}{8})+\dfrac{3}{16}(\dfrac{1}{8})=\dfrac{7}{64}$

$P(X=12)=\dfrac{1}{8}(\dfrac{3}{16})+3(\dfrac{1}{8})(\dfrac{1}{8})+\dfrac{3}{16}(\dfrac{1}{8})=\dfrac{3}{32}$

$P(X=13)=\dfrac{1}{8}(\dfrac{3}{16})+2(\dfrac{1}{8})(\dfrac{1}{8})+\dfrac{3}{16}(\dfrac{1}{8})=\dfrac{5}{64}$

$P(X=14)=\dfrac{1}{8}(\dfrac{3}{16})+(\dfrac{1}{8})(\dfrac{1}{8})+\dfrac{3}{16}(\dfrac{1}{8})=\dfrac{1}{16}$

$P(X=15)=\dfrac{1}{8}(\dfrac{3}{16})+\dfrac{3}{16}(\dfrac{1}{8})=\dfrac{3}{64}$

$P(X=16)=\dfrac{3}{16}(\dfrac{3}{16})=\dfrac{9}{256}$

a. Probability distribution of x

$\def\arraystretch{1.5} \begin{array}{c:c} x & p(x) \\ \hline 2& 1/256\\ \hdashline 3& 1/64\\ \hdashline 4& 1/32\\ \hdashline 5& 3/64\\ \hdashline 6& 1/16\\ \hdashline 7& 5/64\\ \hdashline 8& 3/32\\ \hdashline 9& 15/128\\ \hdashline 10 &1/8\\ \hdashline 11& 7/64\\ \hdashline 12& 3/32\\ \hdashline 13& 5/64\\ \hdashline 14& 1/16\\ \hdashline 15& 3/64\\ \hdashline 16& 9/256\\ \hdashline \end{array}$

b.

$mean=E(X)=2(\dfrac{1}{256})+3(\dfrac{1}{64})+4(\dfrac{1}{32})$

$+5(\dfrac{3}{64})+6(\dfrac{1}{16})+7(\dfrac{5}{64})+8(\dfrac{3}{32})$

$+9(\dfrac{15}{128})+10(\dfrac{1}{8})+11(\dfrac{7}{64})+12(\dfrac{3}{32})$

$+13(\dfrac{5}{64})+14(\dfrac{1}{16})+15(\dfrac{3}{64})+16(\dfrac{1}{256})$

$=9.375$

c. Variance of x

$mean=E(X^2 )=2^2(\dfrac{1}{256})+3^2(\dfrac{1}{64})+4^2(\dfrac{1}{32})$

$+5^2(\dfrac{3}{64})+6^2(\dfrac{1}{16})+7^2(\dfrac{5}{64})+8^2(\dfrac{3}{32})$

$+9^2(\dfrac{15}{128})+10^2(\dfrac{1}{8})+11^2(\dfrac{7}{64})+12^2(\dfrac{3}{32})$

$+13^2(\dfrac{5}{64})+14^2(\dfrac{1}{16})+15^2(\dfrac{3}{64})+16^2(\dfrac{1}{256})$

$=99.6328125$

$Var(X)=\sigma^2=E(X^2)-(E(X))^2$

$=99.6328125-(9.375)^2=11.7421875$