Answer:
The required probability is:
[tex]\dfrac{1}{9}[/tex]
Step-by-step explanation:
It is given that Emma has two fair dice.
To win, she needs to score 5 using both dice.
Cases, where she can have 5:
{(1,4), (4,1), (2,3), (3,2)}
Let the event be A that she gets a score = 5
Total number of favorable cases, n(A) = 4
Total number of cases for throw of 2 dice, n(S) = 36
These are:
{
(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)
}
Formula for probability of an event E is given as:
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]
or
[tex]P(E) = \dfrac{\text{n(E)}}{\text {n(S)}}[/tex]
Using the above formula, the required probability:
[tex]P(A) = \dfrac{\text{n(A)}}{\text {n(S)}}\\\Rightarrow \dfrac{4}{36}\\\Rightarrow \dfrac{1}{9}[/tex]
So, the required probability is [tex]\dfrac{1}{9}[/tex].