Probability Problem

#1
In a group of 50 boys, 18 own a laptop, 20 own a desktop, and of these 6 own both. What is the probability that a boy picked from this group at random owns either a laptop or a desktop computer, but not both.

I answered 13 out of 25 but was told I am wrong. I was told the right answer is 16 out of 25 but I disagree, and they could not explain why 16 was right. Can anyone illuminate this answer for me? Thanks in advance,

Cole
 
#2
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.)
 
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