The quickest way to solve this question is indeed to simply compute the probability that X gets the 1st slot (\(\frac{1}{15}\)) and the probability that X gets the 2nd slot (again \(\frac{1}{15}\)) and then add the two.

However, it may be useful for at least a few students to know another method to solve this problem:

Attachment:

Probability.PNG [ 2.83 KiB | Viewed 1619 times ]
The number of ways in which the first ad slot can be filled = 15

Number of ways in which the second ad slot can be filled (after the first one has already been filled) = 14

So, the number of ways in which both the slots can be filled together = 15*14

Now, let's consider the case when X gets Slot 1:

Attachment:

Probability2.PNG [ 2.22 KiB | Viewed 1617 times ]
The number of ways now in which the second slot can be filled = 14.

So, total number of ways in which X can get Slot 1 = 14

Similarly, the total number of ways in which X can get Slot 2 = 14

Probability of X getting Slot 1 or Slot 2 = (Number of ways in which X can get Slot 1 or Slot 2)/ (Total number of ways in which the 2 slots can be allocated)

=\(\frac{(14+14)}{(15*14)}\)

=\(\frac{2}{15}\)

_________________

Please press Kudos if you were helped by my post!