The finalists in an essay competition are Lisa (L), Melina (M), Ben (B), Danny (D), Eric (E), and Joan $(J)$. Consider these finalists to be a population of interest. The possible samples (without replacement) of size two that can be obtained from this population of sib finalists are as follows. L.M L.B L,D L,E L,J M,B M,D $M, E M, J$ B,D B,E B,J D.E D.J E,J If a simple random sampling method is used to obtain a sample of two of the finalists, what are the chances of selecting Lisa and Danny? A. $\frac{1}{6}$ B. $\frac{1}{15}$ C. $\frac{1}{3}$ D. $\frac{2}{15}$

Step 1 ：The problem is asking for the probability of selecting Lisa and Danny from a group of 6 people. This is a combination problem, as the order of selection does not matter.

Step 2 ：The total number of ways to select 2 people from 6 is given by the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\), where n is the total number of items, k is the number of items to choose, and '!' denotes factorial. In this case, n=6 and k=2.

Step 3 ：Calculating the total combinations, we get 15.

Step 4 ：The event of interest is selecting Lisa and Danny, which is one specific combination. Therefore, the number of favorable outcomes is 1.

Step 5 ：The probability is the ratio of the number of favorable outcomes to the total number of outcomes. So, the probability is \(\frac{1}{15}\).

Step 6 ：Final Answer: \(\boxed{\frac{1}{15}}\)