Answer to Question #150149 in Statistics and Probability for Boosie

Question

If two dice are rolled one time, find the probability of getting these results.
a. A sum of 9
b. sum of 7 or 11
c. Doubles
d. A sum less than 9
e. A sum greater than or equal to 10

Answer 1

When two dice are thrown simultaneously, thus number of event can be 6² = 36 because each die has 1 to 6 number on its faces.

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

a. A sum of 9

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

$$P(sum=9)=\dfrac{4}{36}=\dfrac{1}{9}$$

b. sum of 7 or 11

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

$$P(sum=9\ or\ sum=11)=\dfrac{8}{36}=\dfrac{2}{9}$$

c. Doubles

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

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

d. A sum less than 9

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

$$P(Sum<9)=\dfrac{26}{36}=\dfrac{13}{18}$$

e. A sum greater than or equal to 10

$$(4,6), (5,5), (5,6), (6,4),(6,5),(6,6)$$

$$P(Sum\geq10)=\dfrac{6}{36}=\dfrac{1}{6}$$