Answer to Question #299585 in Statistics and Probability for rk001

Question

A pair of fair dice is rolled. Let X be the random variable representing the sum of the numbers that appear.

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

b. Find P (X≥ 8).

c. Find P (X≤ 7).

d. Find the probability that X takes an even value.

e. Find P (3 ≤X≤ 10)

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.

$P(3\le X\le 10)=P(X=3)+P(X=4)$

$+P(X=5)+P(X=6)+P(X=7)$

$+P(X=8)+P(X=9)+P(X=10)$

$=\dfrac{2}{36}+\dfrac{3}{36}+\dfrac{4}{36}+\dfrac{5}{36}+\dfrac{6}{36}+\dfrac{5}{36}+\dfrac{4}{36}+\dfrac{3}{36}$

$=\dfrac{8}{9}$