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**ANDS!** The entire space consists of all possible combinations of choosing three questions out of 100. There are however, 90C3 and 90C2*10 possible combinations that will allow him to pass the test so:

Total # of possible 3-question combinations: 100 choose 3

Total # of possible 3-question combinations that student knows all answers: 90C3

Total # of possible 3-question combinations that students knows two answers and doesn't know the last one: (90 choose 2)X10

Probability he will be given a 3-question combinaiton he knows: $\displaystyle \frac{90C3+90C2*10}{100C3}$

This assumes that the test is premade, and its simply a matter of selecting one of the premade tests. If he is actually picking questions then its much easier:

$\displaystyle \frac{90}{100}\frac{89}{99}+\frac{90}{100}\frac{89 }{99}\frac{88}{98}$