# Urn problem with intermediate step

Suppose there is an urn containing 120 black and 30 white balls. Now, you randomly draw 100 balls (without replacement). Then, you draw 2 balls again (without replacement) from the selected 100 balls. What is the probability that both balls drawn are white? Intuitively, I would say that the intermediate step of selecting the 100 balls should not matter. Thus, I would calculate it as $$\frac{30}{150} \times \frac{29}{149}$$. My question would be: How can one formally model the intermediate step with the 100 balls and show that it does not matter?

• Yes there is a probability that you extract 100 black ball Commented Apr 20 at 10:47

Your intuition that the intermediate step doesn't matter, turns out to be correct, although this certainly wasn't obvious to me.

Let $$\ W_1\$$ be the number of white balls in the $$100$$ first drawn, and $$\ W_2\$$ the number of white balls in the $$2$$ drawn second. By the law of total probability, the quantity you have to calculate is $$\Bbb{P}\big(W_2=2\big)=\sum_{w=2}^{30}\Bbb{P}\big(W_2=2\,\big|\,W_1=w\big)\Bbb{P}\big(W_1=w\big)\ .\tag{1}\label{e1}$$ Now if there are $$\ w\$$ white balls in the $$100$$ from which the final two are drawn, then the probability that both of those will be white is $$\ \frac{w(w-1)}{100\cdot99}=\,\Bbb{P}\big(W_2=2\,\big|\,W_1=w\big)\ .$$

There are $$\ {30\choose w}\$$ ways of choosing $$\ w\$$ white balls from the $$\ 30\$$ initially in the urn and for each of those there are $$\ {120\choose100-w}\$$ of choosing the remaining $$\ 100-w\$$ black balls from the $$120$$ initially in the urn. There are thus $$\ {30\choose w}{120\choose100-w}\$$ ways in which $$\ w\$$ white and $$\ 100-w\$$ black balls can be drawn from the $$150$$ balls initially in the urn. Since there are $$\ {150\choose100}\$$ equally likely ways in which $$100$$ balls can be chosen from the $$150,$$ we have $$\ \Bbb{P}\big(W_1=w\big)=\frac{{30\choose w}{120\choose100-w}}{{150\choose100}}\ .$$ Substituting these values into equation (\ref{e1}) gives \begin{align} \Bbb{P}\big(W_2=2\big)&=\sum_{w=2}^{30}\frac{w(w-1)}{100\cdot99}\frac{{30\choose w}{120\choose100-w}}{{150\choose100}}\\ &=\sum_{w=2}^{30}\frac{w(w-1)\,30!\,120!\,100!\,50!}{100\cdot99\cdot w!\,(30-w)!\,(100-w)!\,(20+w)!\,150!}\\ &=\sum_{w=2}^{30}\frac{98!\,50!\,30!\,120!}{(w-2)!\,(100-w)!\,(30-w)!\,(20+w)!\,150!}\\ &=\sum_{w=2}^{30}{98\choose w-2}{50\choose20+w}{150\choose30}^{-1}\\ &=\frac{\sum_\limits{v=0}^{28}{98\choose v}{50\choose22+v}}{{150\choose30}}\\ &=\frac{\sum_\limits{v=0}^{28}{98\choose v}{50\choose28-v}}{{150\choose30}}\\ &=\frac{{148\choose 28}}{{150\choose30}}\\ &=\frac{30\cdot29}{150\cdot149}\ . \end{align} The identity $$\ \frac{\sum_\limits{v=0}^{28}{98\choose v}{50\choose22+v}}{{150\choose30}}={148\choose 28}\$$ used in the penultimate step above is an instance of Vandermonde's identity.

• There is a chance that you extract only 100 black ball, so w = 0, i thought thats why its important how we extract the balls Commented Apr 20 at 21:22

First shuffle the balls and place them in a line. We are interested in the sequences which start with two whites, and there are

$$\binom{148}{28}$$

of them. As there are $$\binom{150}{30}$$ such sequences, the probability is

$$\frac{30\cdot29}{150\cdot149}$$

due to

$$k^{\underline a}\binom{n}{k}=n^{\underline a}\binom{n-a}{k-a}$$

If we truncate the sequences to length $$100$$, this will not affect the number of sequences, and therefore nor the probability of $$2$$ whites at the start.