Answer to Question #231203 in Statistics and Probability for Nelson

Question

If two dice are rolled one time. Find the probability of getting the results. (A) A sum of 6 ( b) doubles ( c) A sum of 7 or 11 ( d) A sum greater than 9 (e) A sum less or equal.

Answer 1

Two dice are rolled one time

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

There are $6^2=36$ outcomes.

a)

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

$$P(sum=6)=\dfrac{5}{36}$$

b)

$$(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$$

$$P(doubles)=\dfrac{6}{36}=\dfrac{1}{6}$$

c)

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

$$5+6=6+5=11$$

$$P(sum=7\ or\ 11)=\dfrac{6}{36}+\dfrac{2}{36}=\dfrac{2}{9}$$

d)

$$4+6=5+5=6+4=10>9$$

$$5+6=6+5=11>9$$

$$6+6=12>9$$

$$P(sum>9)=\dfrac{3}{36}+\dfrac{2}{36}+\dfrac{1}{36}=\dfrac{1}{6}$$

e)

$$1+1=2\leq4$$

$$1+2=2+1=3\leq4$$

$$1+3=2+2=3+1=4\leq4$$

$$P(sum\leq4)=\dfrac{1}{36}+\dfrac{2}{36}+\dfrac{3}{36}=\dfrac{1}{6}$$