A box contains 2 red chips and 3 black chips. If chips are randomly removed from the box one at a time, without replacement, until all of the chips are removed, what is the probability that the last two chips are black?

Let the 5 chips be R1,R2,B1,B2,B3

The event is a success when the last two chips are black which is possible in the below combinations,
R1,R2,B1,B2,B3
R1,R2,B2,B1,B3
R1,R2,B3,B1,B2

In each of the above combinations the black chips at the last can be arranged in 2! i.e 2 ways and the first three chips can be arranged in 3! i.e. 6 ways
thus the total number of ways to obtain black chips at the last two turns is 3! X 2! X3 = 36

Total number of ways in which five chips can be selected is 5! ways = 120

Thus the probability of the events in which the last two turns will yield black chips is 36/120 = 3/10

Hence option C is the answer