Answer to Question #295284 in Statistics and Probability for Mina

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

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

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

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

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

$$=99.6328125$$

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

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