Question: Three numbers are chosen at random without replacement from $(1, 2, 3 \dots, 10)$. The probability that the minimum number is $3$ or the maximum number is $7$.
My logic: since the question has specified no replacement what I did was first fix $3$ in one place then pick out other no.s from remaining $7$ and again fixed $7$ and then took out other no.s.$(1/10)(7/9)6/8 + (1/10)(6/9)5/8 = 1/10 $, but the answer is 11/40. Please tell me where am I going wrong.
There are ${7 \choose 2}=21$ sets with $3$ the minimum and ${6 \choose 2}=15$ sets with $7$ the maximum out of ${10 \choose 3}=120$ total. We have counted three sets $(347,357,367)$ twice, so the total probability is $$\frac {21+15-3}{120}=\frac {11}{40}$$