Probability problem without replacement and series I have a probability problem but I am unsure about the result.
I would like to confirm my guesses.
During an exam, 10 students are drawing one by one a question (among 12 questions) without putting the question back.
One student did not study 3 questions.
Let x be the number of passage of the student, and p the probability that he draws a question he did not study.
1) Is p independent of x.
2) Is p a decreasing function of x (ie, the more the student waits, the less likely he will draw a question he did not study).
For me:
1) clearly, $p$ is dependent of $x$.
2) My guess is that it has to do with the total probability formula, to get a "sequence", and then, we have to tell if the sequence is increasing or decreasing. 
if $x=1$, then $p=3/12$.
if $x=2$, then it depends on what the first student drew.
Let Bi: we draw a "bad" question on the i-th draw.
Let Gi: we draw a "good" question on the i-th draw
We can have either: B1 & B2 or G1 & B2.
So using the total probability formula, we have:
P(B2)= 3/12 * 2/11 + 9/12 * 3/11 = 3/12
Can somebody help to find the more general formula please ?
Fixed: figures in my formula.
Thanks.
 A: The probability the student will draw a question he did not study is $3/12$, no matter when the student draws the question. This is because all permutations of the questions are equally likely. The probability that the question at position $8$ is bad is just the same as the probability that the question at position $1$, or $9$, is bad.
If you have $3$ black balls, and $9$ red, and arrange them at random, any position is just as likely to be occupied by a black ball as any other position.
Remark: If the above is not obvious, let us calculate. It is clear that if the student is first, his probability of drawing a bad question is $\frac{3}{12}$.
Let us do a painful cases analysis of the probability the student draws a bad question if he is second.
This could happen in two ways: (i) First question is bad, and second question is bad or (ii) First question is good and second is bad.
We first find the probability of (i). The probability first is bad is $\frac{3}{12}$. Given that, the probability second is bad is $\frac{2}{11}$, for a probability of $\frac{3}{12}\cdot \frac{2}{11}$.
Similarly, the probability of (ii) is $\frac{9}{12}\cdot\frac{3}{11}$.
Add. We get $\frac{3}{12}\cdot\frac{2}{11}+\frac{9}{12}\frac{3}{11}$. Simplify. We get $\frac{3}{12}$, the same answer as the one we got the "easy" way above.
Now if you are in a mood to do unpleasant work, you could find  the hard way the probability that the student draws a bad question, given that he is third in line. This can happen in several ways: GGB, GBB, BGB, BBB. Calculate all these probabilities, add them. Again you will get $\frac{3}{12}$. 
After that, it is unlikely you will want to calculate the long way the probability given the student is fourth in line! 
