Question 1119808
This would be 6C2*9C5, the number of ways 6 can have 2 with brake problems * the number of ways the other 9 would have 5 brake problems.
Denominator is 15C7 (the sum of the 6 and 9 and the 2 and 5.

This is 1890/6435 or 0.2937
Another way is to take a sample of 6 with 2
7/15*6/14*8/13*7/12*6/11*5/10=70560/(15*14*13*12*11*10)=0.0196. There are 15 of these ways where 2 will have brake problems and  where the numerator is ordered differently but with the same numbers, so all 15 is 0.2937.