I think you should divide this problem into several cases.

Exactly 1

C(10,1)*C(7,3)/C(17,4)

Exactly 2

C(10,2)*C(7,2)/C(17,4)

Exactly 3

C(10,3)*C(7,1)/C(17,4)

Exactly 4

C(10,4)*C(7,0)/C(17,4)

And add them up to get your answer.

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There is an easier way.

Since we want either 1,2,3, or all 4.

Find the probably of its negation, that is, 0.

Hence,

C(10,0)*C(7,4)/C(17,4)

Then 1 (total probability) subtract this result.

To get the same answer.