Question #150149
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
1
Expert's answer
2020-12-11T13:10:14-0500

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.


1234561(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)2(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)3(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)4(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)5(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)6(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\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)(3,6), (4,5), (5,4), (6,3)

P(sum=9)=436=19P(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)(1,6),(2,5),(3,4), (4,3), (5,2),(5,6), (6,1),(6,5)

P(sum=9 or sum=11)=836=29P(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)(1,1), (2,2), (3,3), (4,4),(5,5),(6,6)

P(doubles)=636=16P(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),(1,1), (1,2), (1,3), (1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(2,1), (2,2), (2,3), (2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,1), (3,2), (3,3), (3,4),(3,5),(4,1),(4,2),(4,3),(4,4),(4,1), (4,2), (4,3), (4,4),(5,1),(5,2),(5,3),(5,1), (5,2), (5,3),(6,1),(6,2)(6,1), (6,2)

P(Sum<9)=2636=1318P(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)(4,6), (5,5), (5,6), (6,4),(6,5),(6,6)

P(Sum10)=636=16P(Sum\geq10)=\dfrac{6}{36}=\dfrac{1}{6}


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