# Finding the probability of selecting a certain ball on the $j$th draw from $M$ balls with $W$ white balls

There are several parts to this question, but I feel that if I understand the first part I might be able to understand the other parts.

The question I am given is: In a box containing $M$ balls, of which $W$ are white, a sample of $n$ balls is drawn without replacement. Let $A_j$, where $j=1,2,...,n$ represent the event that the ball drawn on the $j$th draw is white. Let $B_k$ denote the event that the sample contains k white balls. Find the probability of $A_j$.

This is what I've understood so far: \begin{align} \text{selecting a white ball first: }P(A_1)&=\frac{W}{M}\\\text{selecting a non-white ball, then a white ball: }P(A_2)&=\frac{M-W}{M}+\frac{W}{M-1}\\\text{selecting two non-white, then a white ball: }P(A_3)&=\frac{M-W}{M}+\frac{M-W-1}{M-1}+\frac{W}{M-2}\end{align} Is this train of thought correct? If it is, I'm not really seeing a way to deduce $P(A_j)$. Instead of this, should I be thinking about $\frac{M!}{(M-n)!}$ because order seems to be important and I'm selecting $n$ balls from the total $M$ balls?

I'm really confused at this point, and any help would be greatly appreciated. I tried to search other probability problems that were similar, but couldn't find anything. Thanks for any hints or links to similar problems in advance!

• The probability is $\frac{W}{M}$ since all sequences of draws are equally likely. – André Nicolas Feb 19 '14 at 19:30
• @AndréNicolas that's it? It's just that simple? – user66807 Feb 19 '14 at 19:35
• Yes, it is that simple. If you want to suffer a bit, calculate the long way the probability the second is white. This is $\frac{W}{M}\cdot \frac{W-1}{M-1}+\frac{M-W}{M}\cdot \frac{W}{M-1}$. Simplify. You will get $\frac{W}{M}$. If you want to suffer more, find the probability the third is white, by considering cases. After some pain the answer will simplify to $\frac{W}{M}$. – André Nicolas Feb 19 '14 at 19:49
• Note that you were not computing, for example, the probability that the second ball is white. You were computing the probability the first is black and the second white. Second is white can also happen via first is white and second is white. – André Nicolas Feb 19 '14 at 19:53
• @AndréNicolas Oh I see what you're doing. I'll suffer the pain and simplify in order to be satisfied. In regards to the probabilities I was computing, I thought since they were asking for $A_j$ that would mean that the ball on the $j$th draw was white implied that the previous balls weren't white, but I guess that's not how it works, right? – user66807 Feb 19 '14 at 20:01

• By multiplying them I am kind of seeing a pattern, mainly the probability of each drawing contains $\frac{W}{M}$. Like this $P(A_1)=\frac{W}{M}$. $P(A_2)=(\frac{W}{M})(\frac{M-W}{M-1})$, and $P(A_3)=(\frac{W}{M})(\frac{M-W}{M-1})(\frac{M-W-1}{M-2})$. Am I still missing the pattern? – user66807 Feb 19 '14 at 19:51
• So would the denominator eventually become $M!$? If so would the numerator work out to be $W(M-W)!$ eventually as well? – user66807 Feb 19 '14 at 20:04