Answer:
Option B is correct.
Explanation:
We have been given that 3 lottery games have decided to play
So, here we have total possibility of numbers s between 0 and 9 which is total of 10 numbers
And number can be chosen more than once.
So, the probability will be
![(1)/(10)\cdot (1)/(10)\cdot (1)/(10)](https://img.qammunity.org/2020/formulas/mathematics/high-school/b8otiy8qcc2t7op6ydmllxgjtv9nrw5395.png)
On simplification we will get
the probability of winning
![(1)/(1000)](https://img.qammunity.org/2020/formulas/mathematics/high-school/9un751imp1lm5hry8287w4jzy2xe0bo814.png)
Therefore, option B is correct.