Answer to Question #305456 in Statistics and Probability for angelica

Question

construct the probability distribution of X for a pair of dice

Answer 1

There are $6^2=36$ outcomes

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 1+1 & 1+2 & 1+3 & 1+4 & 1+5 & 1+6 \\ \hdashline 2 & 2+1 & 2+2 & 2+3 & 2+4 & 2+5 & 2+6 \\ \hdashline 3 & 3+1 & 3+2 & 3+3 & 3+4 & 3+5 & 3+6 \\ \hdashline 4 & 4+1 & 4+2 & 4+3 & 4+4 & 4+5 & 4+6 \\ \hdashline 5 & 5+1 & 5+2 & 5+3 & 5+4 & 5+5 & 5+6 \\ \hdashline 6 & 6+1 & 6+2 & 6+3 & 6+4 & 6+5 & 6+6 \\ \hdashline \end{array}$$

a.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & x & f & P(X=x) \\ \hline & 2 & 1 & 1/36 \\ \hdashline & 3 & 2 & 2/36 \\ \hdashline & 4 & 3 & 3/36 \\ \hdashline & 5 & 4 & 4/36 \\ \hdashline & 6 & 5 & 5/36 \\ \hdashline & 7 & 6 & 6/36 \\ \hdashline & 8 & 5 & 5/36 \\ \hdashline & 9 & 4 & 4/36 \\ \hdashline & 10 & 3 & 3/36 \\ \hdashline & 11 & 2 & 2/36 \\ \hdashline & 12 & 1 & 1/36 \\ \hdashline Sum= & & 36 & 1 \\ \hdashline \end{array}$$

Construct the probability distribution of X for a pair of dice.

$$\def\arraystretch{1.5} \begin{array}{c:c} x & p(x)\\ \hline 2 & 1/36 \\ 3 & 1/18 \\ 4 & 1/12 \\ 5 & 1/9 \\ 6 & 5/36 \\ 7 & 1/6 \\ 8 & 5/36 \\ 9 & 1/9 \\ 10 & 1/12 \\ 11 & 1/18 \\ 12 & 1/36 \\ \end{array}$$

b.

$$P(X\ge 8)=P(X=8)+P(X=9)+P(X=10)$$ $$+P(X=11)+P(X=12)$$ $$=\dfrac{5}{36}+\dfrac{4}{36}+\dfrac{3}{36}+\dfrac{2}{36}+\dfrac{1}{36}=\dfrac{5}{12}$$

c.

$$P(X\le 7)=P(X=2)+P(X=3)+P(X=4)$$ $$+P(X=5)+P(X=6)+P(X=7)$$ $$=\dfrac{1}{36}+\dfrac{2}{36}+\dfrac{3}{36}+\dfrac{4}{36}+\dfrac{5}{36}+\dfrac{6}{36}=\dfrac{7}{12}$$

d.

$$P(X\ is\ even)=P(X=2)+P(X=4)$$ $$+P(X=6)+P(X=8)+P(X=10)$$ $$+P(X=12)=\dfrac{1}{36}+\dfrac{3}{36}+\dfrac{5}{36}+\dfrac{5}{36}+\dfrac{3}{36}$$ $$+\dfrac{1}{36}=\dfrac{1}{2}$$

e.

$$P(3\le X\le 10)=P(X=3)+P(X=4)$$ $$+P(X=5)+P(X=6)+P(X=7)$$ $$+P(X=8)+P(X=9)+P(X=10)$$ $$=\dfrac{2}{36}+\dfrac{3}{36}+\dfrac{4}{36}+\dfrac{5}{36}+\dfrac{6}{36}+\dfrac{5}{36}+\dfrac{4}{36}+\dfrac{3}{36}$$ $$=\dfrac{8}{9}$$