Let S be the set of permutations of (1,2,3,4,5,6) whose first term is a prime.
If we choose a permutation at random from S, what is the probability that the third term is equal to 2?
The first term can be either 2, 3, or 5.
If the first term is 2, there are 5! = 5 x 4 x 3 x 2 x 1 = 120 ways to permutate the other terms.
Similarly, there are 120 ways if the first term is a 3 and there are 120 ways if the first term is a 5.
Therefore there is a total of 360 possible permutations for the denominator of the probability.
If the third term is a 2 and the first term is a 3, there are 4 x 3 x 2 x 1 = 24 ways.
Similarly, if the third term is a 2 and the first term is a 5, there are 24 ways.
Adding, we get 48 ways for the numerator of the probability.
The probability will be 24 / 360.