# How to solve this urn problem in without replacement model.

An urn contains $$a$$ white balls and $$b$$ black balls. Balls are drawn randomly from the urn without replacement. Find the probability that a white ball is drawn at the $$k$$th draw.

In this question I have tried to put binomial theorem at $$k^{th}$$ instant. As without replacement is given so at $$k^{th}$$ instant or before $$k-1$$ instant there is a chance of white ball appear. So how to proceed?

• This looks like it should use [probability], not [probability-theory]. Commented Nov 26, 2021 at 15:36
• Each draw is the same, each ball is equally likely to be in any given position. Thus, $\frac {w}{w+b}$.
– lulu
Commented Nov 26, 2021 at 15:40
• The binomial theorem: $(x+y)^n=\sum_{r=0}^n \binom nr x^r y^{n-r}.$ The binomial distribution is closely related to this, but applies to drawing with replacement. The only part of the theorem that seems applicable to the question is the notation $\binom nr$; is that what you meant by "binomial theorem"? Commented Nov 26, 2021 at 23:12

I think this problem is probably easiest to solve by applying combinatorics to the order of ball selection. Even if we stop after the $$k$$th draw, pretend we draw out all the balls, and line them up in the order they are drawn. Now there are $$\binom{a+b}{a}$$ ways to arrange the white balls in these orders, and $$\binom{a+b-1}{a-1}$$ ways to arrange them such that a white ball is in the $$k$$th position. Then the probability of a white ball being drawn on the $$k$$th draw is:

$$P=\frac{\binom{a+b-1}{a-1}}{\binom{a+b}{a}}=\frac{\frac{(a+b-1)!}{(a-1)!b!}}{\frac{(a+b)!}{a!b!}}=\frac{a}{a+b}$$

As @lulu mentions, this answer makes sense because the $$k$$th draw should be no different from the first draw.

• Here given is without replacement so your is wrong answer Commented Nov 26, 2021 at 17:12
• @SudhirKumarSahoo My answer fundamentally assumes that balls are drawn without replacement. If there is some more specific issue with my interpretation of the question, please edit your question to make that clear. Commented Nov 26, 2021 at 17:46
• Look at $K^{th}$ drawn all white ball will come means $K=a$ so this is telling that your answer is wrong the way you are doing the process. Commented Nov 27, 2021 at 5:23
• @SudhirKumarSahoo Angelica's solution is correct. You are choosing a sequence of balls from the urn. Imagine you arrange the balls you select in a row until all $a + b$ balls are in the row. Of these balls, $a$ are red. Since the selection of balls from the urn is random, a red ball is equally likely to be anywhere in the row. In particular, the probability that a red ball is in the $k$th position is $\frac{a}{a + b}$, as Angelica found. Commented Nov 27, 2021 at 10:22

The correct answer I found is upto $$K-1$$ lets say all black ball we pick having probability is $$P = \left(\frac{b}{b+w}\right)\left(\frac{b-1}{b+w-1}\right)\left(\frac{b-2}{b+w-2}\right)\left(\frac{b-3}{b+w-3}\right) \cdots..\left(\frac{b-k+1}{b+w-k+1}\right)$$ then final answer is $$1-P$$

• This seems to be the answer to the question, "what is the probability that at least one white ball is drawn before drawing the $k$th ball?". Is that your question? You certainly can use the binomial theorem to count the number of ways you reach $k-1$ draws with $i$ white balls left. But you would need to calculate this for all $i$, and your answer $P$ is the relevant probability just for $i=a$. It's possible, but much harder than the solution I laid out. Commented Nov 26, 2021 at 17:43
• According to what you wrote, the requirement of the question is, "Find the probability that a white ball is drawn at the $k$th draw." Supposing $a = b = 5$ and $k = 4,$ here is an example of "a white ball is drawn at the $k$th draw": $BBBW.$ Here are other examples of "a white ball is drawn at the $k$th draw": $BBWW,$ $BWBW.$ Examples of "a white ball is not drawn at the $k$th draw": $BBBB,$ $BWBB,$ $WWWB.$ Your answer seems to have very little to do with any of this. It does not even give the probability that a white ball is drawn for the first time on the $k$th draw. Commented Nov 26, 2021 at 21:45