Probability that exactly 2 of 3 objects are in 1 of 3 baskets with sizes 5, 8, 2

Question

I want to calculate the probability that some mutation occurs on a certain DNA section by a given number mutations. I rephrased it into this problem:

Three (identical) persons enter a train (section A has $5$ seats, section B has $8$ seats and section C has $2$ seats). What is the probability that exactly (not more or less) $2$ persons sit in section A? (all seats have the same probability)

Answer 1

Let's call the persons 1,2,3 and assume they enter in that order. The case you want to look at are 1-2 or 1-3 or 2-3 in section A while 3 in B or C, 2 in B or C, 1 in B or C.

It results that you don't even need to differentiate between B and C, so let's call it BC with 10 seats.

• Case 1-2 in A, 3 in BC: 5*4*10 (1 has 5 seats, B has 4 seats available, C 10 seats)
• Case 1-3 and 2-3 are identical

You have the $15 \times 14 \times 13 $ possible arrangements to begin with so the final answer if I did not make any silly mistake is $\frac{3\times5\times4\times10}{15 \times 14 \times 13 }$

Answer 2

@Matthias M.:

Think of the seats like this: _ _ _ _ _ , _ _ _ _ _ _ _ _ _ _

You need 2 in the first five seats $\binom{5}{2}=10$

You have one left for the remaining 10 seats $\binom{10}{1}=10$

$$ \frac{\binom{5}{2}\binom{10}{1}}{\binom{15}{3}}=\frac{20}{91}$$