Answer :
Answer:
(9)[tex]\frac{1}{12}[/tex] (10) [tex]\frac{1}{12}[/tex] (11)[tex]\frac{5}{12}[/tex] (12)[tex]\frac{1}{4}[/tex] (13)[tex]\frac{1}{6}[/tex] 14)[tex]\frac{5}{36}[/tex] (15)[tex]\frac{1}{12}[/tex] (16)0
Step-by-step explanation:
The sample Space of the single die rolled twice is presented below:
{(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), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6),}.
n(S)=36
(9)Probability of getting two numbers whose sum is 9.
The possible outcomes are: (3, 6), (4, 5), (5, 4)
[tex]P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}[/tex]
10) Probability of getting two numbers whose sum is 4.
The possible outcomes are: (1, 3),(2, 2),(3, 1),
[tex]P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}[/tex]
11.)Find the probability of getting two numbers whose sum is less than 7.
The possible outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)
[tex]P(\text{two numbers whose sum is less than 7})=\frac{15}{36}=\frac{5}{12}[/tex]
12.Probability of getting two numbers whose sum is greater than 8
The possible outcomes are:(4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)
[tex]P(\text{two numbers whose sum is greater than 8})=\frac{9}{36}=\frac{1}{4}[/tex]
(13)Probability of getting two numbers that are the same (doubles).
The possible outcomes are:(1, 1)(2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
[tex]P(\text{two numbers that are the same})=\frac{6}{36}=\frac{1}{6}[/tex]
14.Probability of getting a sum of 7 given that one of the numbers is odd.
The possible outcomes are: (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
[tex]P(\text{getting a sum of 7 given that one of the numbers is odd.})=\frac{5}{36}[/tex]
(15)Probability of getting a sum of eight given that both numbers are even numbers.
The possible outcomes are: (2, 6), (4, 4), (6, 2)
[tex]P(\text{getting a sum of eight given that both numbers are even numbers.})=\frac{3}{36}\\=\frac{1}{12}[/tex]
16.Probability of getting two numbers with a sum of 14.
[tex]P(\text{getting two numbers with a sum of 14.})=\frac{0}{36}=0[/tex]