Question 1067876
First, we have to know how many cases there are that he passes 2 test, and fails the last test. 

Case 1: pass, pass, fail
Case 2: pass, fail, pass
Case 3: fail, pass, pass

So there are 3 cases. 

Then we have to take into account of the case when he passes all three:

Case 4: pass, pass, pass

Now, you replace each of these scenarios with the probability. 

Case 1: (2/3)*(2/3)*(1/3) = 4/27
Case 2: (2/3)*(1/3)*(2/3) = 4/27
Case 3: (1/3)*(2/3)*(2/3) = 4/27
Case 4: (2/3)*(2/3)*(2/3) = 8/27

Then you add up the probabilities of each of these cases happening:
(4/27)+(4/27)+(4/27)+(8/27) = 20/27
Therefore, the probability that he passes two tests is 20/27.