There are 15 CD's out of which 4 are defective . CD's are selected one by one without replacement(10 times). What is the probability that the last defective ball is chosen at the 10th time?
I have used the following approach
First fix one defective CD for the 10 th choose and fix the 1st , 2nd ,3rd choose for defective CD's ,
Now the probability of this can be chosen by (4/15)(3/14)(2/13)(11/12)(10/11)(9/10)(8/9)(7/8)(6/7)*(1/6)==P(say)...
Now the first 3 defective items can be arranged in (C(9,3) ways and each defective items can go to 10 th place at a time(4 ways) ...
So the required probability is P*C(9,3)*4...
But the Answer is not correct !!!
Hint
A simple way is to imagine all the CDs in a row, and see that if the $4^{th}$ defective is in the ${10}^{th}$ position, the other three must be among the first $9$, so $Pr = \dfrac{\binom93}{\binom{15}4}$
The approach you started with can also lead to the correct answer, now try to figure out what needs to be corrected, and why...