This seems to be a straight foward hypergeometric dist.

I assume we cannot pick a number twice, i.e., we are sampling without replacement.

If you replace the balls it's binomial.

In this case it's

P(X=x)= (6 choose x) (43 choose 6-x)/(49 choose 6)

x=0,1,2,...,6

You just want 4,5 and 6.