Urn $U_i \;\;(i = 1, 2)$ contains $N_i$ balls out of which $r_i$ are red and $N_i − r_i$ are black. A sample of $n \;(1 ≤ n ≤ N_1)$ balls is chosen at random (without replacement) from urn $U_1$ and all the balls in the selected sample are transferred to urn $U_2$. After the transfer two balls are drawn at random from the urn $U_2$. Find the probability that both the balls drawn from urn $U_2$ are red.

My attempt:

Let $Y $ be the number of red balls transferred from $U_1$ to $U_2$

Support of $Y $ is $S_Y =\{0,1,2\ldots r_1\}$ with pmf

$$f_Y (k) = \frac{{r_1\choose k}{N_1-r_1\choose n-k}}{{N_1\choose n}} \; \text{for } k\in S_Y \text{ and } 0 \text{ otherwise}$$

Let $Z$ be the number of balls drawn out of $U_2$

$S_Z = \{0,1,2\} $ with the pmf

$$f_Z(k) = \frac{{r_2 +Y\choose k}{N_2 + n -r_2 -Y\choose 2-k}}{{N_2+n\choose 2}} \; \text{for } k\in S_Z \text{ and } 0 \text{ otherwise} $$ Then, the probability of drawing both red balls from $U_2 $ is


$$ =\sum_{i\;=\;0}^{r_1} Pr(Z=2 |Y = i)P(Y=i) $$

$$ =\sum_{i\;=\;0}^{r_1}\frac{{r_2 + i\choose 2}}{{N_2+n\choose 2}} \frac{{r_1\choose i}{N_1-r_1\choose n-i}}{{N_1\choose n}} $$

Is this correct?

However, my instructor has given the following answer:

$$\frac1{(N_2+1)(N_2 +n-1)} \big[r_2(r_2-1) + 2r_2\frac{n_1}{N_1} + n(n-1)\frac{r_1(r_1-1)}{N_1(N_1-1)}\big]$$

I can't see how to reduce my answer to this above form.


1 Answer 1


First of all I will try to reconstruct he formula given by your instructor, as surely the expression cited in the question cannot be correct.

I would assume the following logic behind the expression.

We can assume that $n$ balls transferred to the second urn are still identifiable. Then the event of draw of one red ball can be written as $$ P(r)=P(1)p(1)+P(2)p(2), $$ where $P(1)=\frac n{N_2+n}$, $P(2)=\frac {N_2}{N_2+n}$ are the probabilities that the ball will be drawn form the 1st or 2nd urn, respectively, and $p(1),p(2)$ are the probabilities to draw a red ball from the respective urn.

For draw of two balls the probabilities will be $$ P(11)=\frac {n(n-1)}{(N_2+n)(N_2+n-1)};\;P(22)=\frac {N_2(N_2-1)}{(N_2+n)(N_2+n-1)};\;\\ P(12)=P(21)=\frac {n N_2}{(N_2+n)(N_2+n-1)};\\ p(11)=\frac{r_1(r_1-1)}{N_1(N_1-1)};\;p(22)=\frac{r_2(r_2-1)}{N_2(N_2-1)};\; p(12)=p(21)=\frac{r_1r_2}{N_1N_2},\, $$ so that the overall probability reads: $$\begin{align} P&=P(11)p(11)+P(22)p(22)+P(12)p(12)+P(21)p(21)\\ &=\frac{1}{(N_2+n)(N_2+n-1)} \left[\frac {n(n-1)}{N_1(N_1-1)}r_1(r_1-1)+r_2(r_2-1)+2\frac n{N_1}r_1r_2\right]. \end{align}$$

I would assume that this is the intended form of the expression given by your instructor and this formula is the one to which your expression can be reduced to, as checked by Mathematica.

  • $\begingroup$ thank you very much. $\endgroup$
    – chesslad
    Jun 21, 2020 at 0:03
  • $\begingroup$ Unfortunately for me, I don't have Mathematica license. Were you able to get to this reduced form from the summation expression using Mathematica? $\endgroup$
    – chesslad
    Jun 21, 2020 at 0:33
  • 1
    $\begingroup$ @Abhay I was able to do this. And already before doing this I had no doubts that your formula is correct. $\endgroup$
    – user
    Jun 21, 2020 at 6:23

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