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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?

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  • $\begingroup$ Yes there is a probability that you extract 100 black ball $\endgroup$
    – HellBoy
    Apr 20 at 10:47

2 Answers 2

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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.

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  • $\begingroup$ 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 $\endgroup$
    – HellBoy
    Apr 20 at 21:22
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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.

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