Question

Problem:
Tad draws three cards at random, without replacement, from a deck of ten cards numbered 1 through 10. What is the probability that no two of the cards drawn have numbers that differ by 1? Express your answer as a common fraction.

Scott's solution:
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there are 9 pairs of cards that differ by one
... 1-2, 2-3, 3-4, etc.

there are 10C2 possible pairs ... 45

so the probability of a pair differing by one is 9/45 or .2
... by more than one is (1 - .2) = .8

3 cards contain 3 pairs
.8^3 = .512

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Thanks a lot Scott, Appreciate your quick response.

I reviewed the answer key and it states 7/15 as the answer to this problem. I am wondering if the answer key is incorrect.