Box $U_1$ contains $1$ white ball and $2$ black balls. Box $U_2$ contains $2$ white balls and $2$ black balls. We extract without reinsertion two balls from every boxes. The four balls are put in a third box $U_3$ initially empty. We randomly extract a ball from $U_3$. Find the probability that the ball is white.

Well, I reasoned in this way. The possible combinations that ensure that $U_3$ contains at least one white ball are BNBB, NBBB, NNBB, BNBN, BNNB, BNNN, NBBN, NBNB, NBNN, NNBN, NNNB. Thus:

  • $\mathbb{P}$($U_3$ contains $3$ white balls)$=\mathbb{P}($(BNBB)$\cap$(NBBB)$)=(\frac{1}{3}\cdot1 \cdot\frac{1}{2}\cdot\frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2}\cdot \frac{1}{3})=0,11$

  • $\mathbb{P}(U_3$ contains $2$ white balls)$=\mathbb{P}($(NNBB)$\cap$(BNBN)$\cap$(BNNB)$\cap$(NBBN)$\cap$(NBNB)$)=(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{1}{3}\cdot 1\cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{1}{3}\cdot 1 \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{3}\cdot \frac{1}{2} \cdot \frac{2}{3})=0,46$

  • $\mathbb{P}(U_3$ contains $1$ white ball)$=\mathbb{P}($(BNNN)$\cap$(NBNN)$\cap$(NNBN)$\cap$(NNNB)$)=(\frac{1}{3}\cdot 1 \cdot \frac{1}{2}\cdot \frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{2}{3})=0,33$

  • $\mathbb{P}(U_3$ doesn't contain any white balls)$=2\mathbb{P}($(NNNN)$)=2(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})=0,1$

Thus $\mathbb{P}($one white ball from $U_3)=\frac{3}{4}\cdot 0,11+\frac{2}{4}\cdot 0,46+\frac{1}{4}\cdot 0,33+\frac{0}{4}\cdot 0,11=0,395$

Is it correct? Particularly I'm interested in reasoning. Thanks in advance.

  • $\begingroup$ You really shouldn't be using decimal numbers in problems like this. It introduces rounding errors, and makes it hard to follow your arithmetic. Use fractions instead. (Also, B is white in Italian, but black in English!) $\endgroup$ – TonyK Jun 30 '20 at 10:43

In the end, it all drills down to selecting one ball. With probability half, it is a ball from $U_1$ and with probability half, it is a ball from $U_2$. The probability of choosing a white ball from each box is known, so the total probability is $\tfrac{1}{2}\cdot\tfrac{1}{3}+\tfrac{1}{2}\cdot\tfrac{1}{2}=\tfrac{5}{12}$.

  • $\begingroup$ Very nice. It makes the other answers look silly. $\endgroup$ – TonyK Jun 30 '20 at 10:46

It does not appear to be correct.

You're drawing two balls without replacement from $U_1$, leaving one behind. The probability that the leftover ball is white (meaning the two balls you drew are black) is $\frac 13$. Thus, the probability that you draw a black ball and a white ball from $U_1$ is $\frac 23$.

You're drawing two balls without replacement from $U_2$, leaving two behind. The probability that both leftovers are black is $\frac 16 = \frac {1}{\binom 42}$, and by symmetry the probability that both leftovers are white also is $\frac 16$. Therefore the probability that you draw a white ball and a black ball from $U_2$ is $\frac 23$.

The draws from $U_1$ and $U_2$ are independent, so we can multiply probabilities.

Probability of no white balls in $U_3$ is $\frac 13 \frac 16 = \frac{1}{18}$.

Probability of one white ball in $U_3$ is $\frac 23 \frac 16+ \frac 13 \frac 23 =\frac 19 + \frac 29 = \frac 13$.

Probability of two white balls in $U_3$ is $\frac 23 \frac 23 + \frac 13 \frac 16= \frac 49 + \frac {1}{18} = \frac 12$.

Probability of three white balls in $U_3$ is $\frac 23 \frac 16=\frac 19$.

Thus, the probability that you draw a white ball is $0 \frac {1}{18}+ \frac 13 \frac 14+ \frac 12 \frac 12 + \frac 19 \frac 34 = \frac {5}{12}$.


Let $X_1, X_2$ be the count of white balls drawn from the first two urn, $Y$ be the sum of these, and $E$ the event that a white ball is then drawn from the third urn.

Your approach was correct. $\mathsf P(E)=\sum_{k=0}^4 \mathsf P(E\mid Y=k)\mathsf P(Y=k)$

However, your evaluations were a bit off.

$$\begin{align}\mathsf P(Y=3)&=\mathsf P(X_1=1)\mathsf P(X_2=2)\\&=\dfrac{\binom 11\binom21}{\binom 32}\dfrac{\binom 22\binom 20}{\binom 42}\\&=\dfrac{1}{9}\\[2ex]\mathsf P(Y=2)&=\mathsf P(X_1=1)\mathsf P(X_2=1)+\mathsf P(X_1=0)\mathsf P(X_2=2)\\&=\dfrac{\binom 11\binom21}{\binom 32}\dfrac{\binom 21\binom 21}{\binom 42}+\dfrac{\binom 10\binom22}{\binom 32}\dfrac{\binom 22\binom 20}{\binom 42}\\&=\dfrac 12\\[2ex]\mathsf P(Y=1)&=\mathsf P(X_1=1)\mathsf P(X_2=0)+\mathsf P(X_1=0)\mathsf P(X_2=1)\\&=\dfrac{\binom 11\binom21}{\binom 32}\dfrac{\binom 20\binom 22}{\binom 42}+\dfrac{\binom 10\binom22}{\binom 32}\dfrac{\binom 21\binom 21}{\binom 42}\\&=\dfrac 13\\[2ex]\mathsf P(Y=0)&=\mathsf P(X_1=0)\mathsf P(X_2=0)\\&=\dfrac{\binom 10\binom22}{\binom 32}\dfrac{\binom 20\binom 22}{\binom 42}\\&=\dfrac 1{18}\end{align}$$


The question can be considered as two events' intersection, that a). $X$ white balls drawn from $U_1$ and $U_2$; b). the ball drawn from $U_3$ is white.

For the probability of event (a), we can consider it as a hypergeometric case. We’ve already known that there’re 3 white balls totally, so X∈(0,1,2,3). Applying the PMF of hypergeometric distribution: $P(X)=\left(\frac {{3 \choose X}{4 \choose 4-X}}{{7\choose 4}}\right)$

In terms of event b, it’s nothing more than $ P(white)$=$X/4$

So what you need to do is: $P=\sum_{X=0}^{4} P(X)P(white)$

  • 1
    $\begingroup$ You should use the conditional probability: $\mathsf P(\text{White}\mid X)=\frac X4$ $\endgroup$ – Graham Kemp Jun 30 '20 at 10:46
  • $\begingroup$ True bro, that’s my bad $\endgroup$ – Harry Lew Jun 30 '20 at 10:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.