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Answer:
[tex]a.\ \frac{5}{36}\\b.\ \frac{1}{12}\\c.\ \frac{2}{9}\\d.\ $Not mutually exclusive$[/tex]
Step-by-step explanation:
Given two fair dice are rolled, one green and one red.
To find:
a. Probability of getting a sum of 8.
b. Probability of getting a sum of 4.
c. Probability of getting a sum of 8 or 4.
d. Are these outcomes mutually exclusive?
Solution:
First of all, let us have a look at the output of roll of two dice.
There are a total of 36 outcomes.
{(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)}
a. Getting a sum equal to 8, we have following outcomes:
{(2, 6), (3,5), (4, 4), (5,3), (6, 2)}
5 possible outcomes for this event.
Formula for probability of an event E can be observed as:
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]
[tex]P($Getting 8$) = \dfrac{5}{36}[/tex]
b. Getting a sum equal to 4, we have following outcomes:
{(1, 3), (2,2), (3, 1)}
3 possible outcomes for this event.
Formula for probability of an event E can be observed as:
[tex]P($Getting 4$) = \dfrac{3}{36} = \dfrac{1}{12}[/tex]
c. Probability of getting a sum of 8 or 4?
Following possible outcomes are there:
{(2, 6), (3,5), (4, 4), (5,3), (6, 2), (1, 3), (2,2), (3, 1)}
Total 8 possible outcomes:
[tex]P($Getting 8 or 4$) = \dfrac{8}{36} = \dfrac{2}{9}[/tex]
d. Are these outcomes mutually exclusive?
These events are not mutually exclusive because they contain the similar outcome from both of the dice.
i.e. (4, 4) and (2, 2)