{1,2,3,5,6,7}
To have a mean of less than 5, the three numbers chosen must have
a sum less than 15.
Let's go for the probability of the complement event, that is,
the choices of three numbers from the set that will have a sum
of 15 or more. Then we will subtract that probability from from 1.
To have a sum that large, the 7 must be chosen, since 6+5+3 is only
14.
If the 7 and 6 are chosen, the third choice could be 2,3, or 5. That's
3 ways.
If the largest two chosen are 7 and 5, then only the 3 could be chosen.
That's 1 more way.
So there are only 3+1 or 4 choices of three that have a sum of 15 or greater.
There are 6C3 = 20 possible choices, so the probability of the sum being
15 or more is 4/20 or 1/5.
Therefore the answer to the given problem is
Edwin