Probability of getting at least 2 numbers smaller of the average of n numbers if you pick 3 of them

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I have got the numbers " 13, 15, 12, 9, 6, 3, 5, 12, 17, 10, 12, 9, 6, 8, 10, 8, 10" and I have to find ,if I randomly pick three of these numbers, what is the probability that at least two of them are smaller than the average of all seventeen?

The average is 9.7 = 10. I have 8 numbers smaller than 10 . The probability of getting at least 2 smaller than the average is P(of getting 2 smaller and 1 greater) + P(of getting 3 smaller and no greater) so is (8/17 * 7/16 *9/15 )+ (8/17 * 7/17 * 6/17 ) . I am not sure if that is the correct way . Can anyone tell me if am right?

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Pretty much! It's just that you can pick those 2 smaller + 1 not smaller in 3 different ways, because you can pick that not smaller one as your first, second, or third. So, you get:

3* (8/17 * 7/16 *9/15 )+ (8/17 * 7/16 * 6/15 )

(also note you had 17 in the denominator for each of the terms in the second expression, which wasn't right, but that was just a typo on your part since you had it right in the first term)