A die with 6 sides is rolled and the number on top is observed. What is the probability of rolling a die once and observing a number less than 5?

1/6

2/3

5/6

4

So P (a 6 in two rolls) = P (6 in first roll) + P (6 in second roll) - P (6 in both rolls)

= 1/6+1/6-1/6*1/6

= 11/36

You could have also used:

P (a 6 in two rolls) = P (6 in the first roll only) + P (6 in the second roll only) + P (6 in both rolls)

= 1/6*5/6 + 5/6*1/6 + 1/6*1/6

= 11/36

The above approach is not easy to handle when the number of rolls are increased.

for two rolls:

for three rolls:

and so on ... Inclusion-exclusion principle

Fortunately there is a better approach in this case, which will be clear if you see the venn diagrams in the link above,

P (a 6 in two rolls) = 1 - P (no 6 in two rolls)

P (no 6 in two rolls) = P (no 6 in first roll) * P (no 6 in second roll)

and P (no 6 in first roll) = P (no 6 in second roll) = 5/6

from there you get:

P (a 6 in two rolls) = 1 - (5/6) ^2

similarly for six rolls it will be 1 - (5/6) ^6 so its 'c'