Seating arrangments

First consider the probability that either gets on one of the ends and then that, out of the remaining people, the other of the pair sits at the other end:

.4*.25=.1

The sample space consists of all possible ways of seating five people in five chairs, 5! = 120.

The event space consists of all possible ways of seating the five people so that Marvin and Juan both occupy end seats. The number of these possibilities can be determined as follows:

Task 1. Put Marvin at the leftmost and Juan at the rightmost then arrange the other 3 people in the remaining 3 seats. This can be done in 3! = 6 ways.

Task 2. Put Juan at the leftmost and Marvin at the rightmost then arrange the other 3 people in the remaining 3 seats. This can be done in 3! = 6 ways.

$P = \frac{6 + 6}{120} = \frac{1}{10} = 0.1$