Three boxes with balls

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.

• 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!) – 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 $$\tfrac{1}{2}\cdot\tfrac{1}{3}+\tfrac{1}{2}\cdot\tfrac{1}{2}=\tfrac{5}{12}$$.

• Very nice. It makes the other answers look silly. – 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: $$P=\sum_{X=0}^{4} P(X)P(white)$$

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