Answer:
[tex]Pr = \frac{3}{36}[/tex]
Step-by-step explanation:
Given
2 number die
Required
P(Sum = 4)
The sample space of 2 die is:
[tex]S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6), (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)[/tex]
[tex](3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6), (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)[/tex]
[tex](5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6), (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)[/tex]
[tex]n(S)= 36[/tex]
The pairs that adds up to 4 are:
[tex]Sum(4) = \{(1,3), (2,2),(3,1)\}[/tex]
[tex]n(Sum) =3[/tex]
So, the probability is:
[tex]Pr = \frac{n(Sum)}{n(S)}[/tex]
[tex]Pr = \frac{3}{36}[/tex]
