Answer:
![(9)/(10)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zq2926wld86uu0cpdt78jx1rdy58b1rgdv.png)
Explanation:
From the table , the total number of numbers = 10
The group of 5 digits contain less than or equal to 2 even digits = 55170
i.e. The total group of 5 digits contain at least 2 =
![10-1=9](https://img.qammunity.org/2020/formulas/mathematics/middle-school/tny4u5na52t8xmk6y157ormtvflkvhean1.png)
Now, the probability that a group of 5 random digits will contain at least 2
even digits is given by :-
![=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}=(9)/(10)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/iyxy1qsi2doct6gnlz3puh58646bq8nmuk.png)
Hence, the probability that a group of 5 random digits will contain at least 2 even digits
![=(9)/(10)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t80v2ede208aomonm5cs6yjti381nlacxm.png)