Answer:
![(1)/(36)](https://img.qammunity.org/2020/formulas/mathematics/college/s4plpocybn3d5gu9r1u6sgglexgsvln6hx.png)
Explanation:
We know that when two events A and B are independent , then the probability of getting A and B will be :-
![P(A\cap B)=P(A)* P(B)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/3lj247zkagi7dw4nry1tpqbzx7yetyhb0h.png)
In a fair dice,total outcomes =6
Number of 2's =1
Then, the probability of getting a 2 =
![(1)/(6)](https://img.qammunity.org/2020/formulas/mathematics/high-school/hpryu5gz6b2hrulak2xp6643a8fps5q09c.png)
If you roll one die two times, then the probability of getting a 2 on the first roll and a 2 on the second roll will be:-
![(1)/(6)*(1)/(6)=(1)/(36)](https://img.qammunity.org/2020/formulas/mathematics/college/p8xl1i5e1h16pvurf2055fs1to8wpuyd5p.png)
Hence, the probability of getting a 2 on the first roll and a 2 on the second roll=
![(1)/(36)](https://img.qammunity.org/2020/formulas/mathematics/college/s4plpocybn3d5gu9r1u6sgglexgsvln6hx.png)