Answer:
![(1)/(6)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/t6mk2mfk5g8w2jv7s83cjb4sbxcq55r7z0.png)
Explanation:
Given that two fair dice are tossed.
Total outcomes as we toss two dice are given as:
{
(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)
}
Where (a, b) represents a is the outcome of the die rolled for the 1st time and b is the outcome of 2nd toss of the die.
Total number of outcomes = 36
The outcomes with sum = 7 are :
{(1, 6), (2, 5), (3, 4) , (4, 3), (5, 2), (1, 6)}
Number of outcomes = 6
Probability of an event E is given as:
![P(E) = \frac{\text{Number of favorable cases}}{\text {Total number of cases}}](https://img.qammunity.org/2021/formulas/mathematics/high-school/aeg3h4h3bbx73banosb6zhsdb88ck3qbng.png)
![P(sum=7) = (6)/(36)\\\Rightarrow \bold{P(sum=7) = (1)/(6)}](https://img.qammunity.org/2021/formulas/mathematics/college/8w65ce3h5g0gktc73bchpp4em4t0dybf4b.png)