Question

A single die is rolled twice. The set of 36 equally likely outcomes is {(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),}. 9. Find the probability of getting two numbers whose sum is 9. 10.Find the probability of getting two numbers whose sum is 4. 11.Find the probability of getting two numbers whose sum is less than 7. 12.Find the probability of getting two numbers whose sum is greater than 8 13.Find the probability of getting two numbers that are the same (doubles). 14.Find the probability of getting a sum of 7 given that one of the numbers is odd. 15.Find the probability of getting a sum of eight given that both numbers are even numbers. 16.Find the probability of getting two numbers with a sum of 14.

Answer 1

Answer:

(9)$\frac{1}{12}$  (10) $\frac{1}{12}$  (11)$\frac{5}{12}$  (12)$\frac{1}{4}$  (13)$\frac{1}{6}$ 14)$\frac{5}{36}$ (15)$\frac{1}{12}$  (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)

$P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}$

10) Probability of getting two numbers whose sum is 4.

The possible outcomes are:  (1, 3),(2, 2),(3, 1),

$P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}$

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)

$P(\text{two numbers whose sum is less than 7})=\frac{15}{36}=\frac{5}{12}$

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)

$P(\text{two numbers whose sum is greater than 8})=\frac{9}{36}=\frac{1}{4}$

(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)

$P(\text{two numbers that are the same})=\frac{6}{36}=\frac{1}{6}$

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)

$P(\text{getting a sum of 7 given that one of the numbers is odd.})=\frac{5}{36}$

(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)

$P(\text{getting a sum of eight given that both numbers are even numbers.})=\frac{3}{36}\\=\frac{1}{12}$

16.Probability of getting two numbers with a sum of 14.

$P(\text{getting two numbers with a sum of 14.})=\frac{0}{36}=0$