I agree with the answer you have given, i.e. P = 13/25. A breakdown of the information provided by the problem reveals that of the 50 boys:

- 12 own a laptop only (Group A);
- 14 own a desktop only (Group B);
- 6 own both (Group C); and
- 18 own neither (Group D).

To meet the required criterion of a boy who owns only one of the computer types, our random selection must come from Group A or Group B, giving a probability P = (12+14)/50 = 26/50 = 13/25.

I suspect that their (wrong) answer comes from an inappropriate application of a formula: P’ = (18+20-6)/50 = 32/50 = 16/25.

(Afterthought: Perhaps the phrase “and of these” is meant to refer only to the “20 [who] own a desktop.” That is, the group breakdown is meant to be:

- 18 own a laptop only (Group A);
- 14 own a desktop only (Group B);
- 6 own both (Group C); and
- 12 own neither (Group D).

In this case, the problem was very badly phrased and the answer is indeed P = (18+14)/50 = 32/50 = 16/25.)