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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.

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  • $\begingroup$ 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
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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 $\tfrac{1}{2}\cdot\tfrac{1}{3}+\tfrac{1}{2}\cdot\tfrac{1}{2}=\tfrac{5}{12}$.

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  • $\begingroup$ Very nice. It makes the other answers look silly. $\endgroup$ – TonyK Jun 30 '20 at 10:46
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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 $0 \frac {1}{18}+ \frac 13 \frac 14+ \frac 12 \frac 12 + \frac 19 \frac 34 = \frac {5}{12}$.

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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.

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

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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: $P=\sum_{X=0}^{4} P(X)P(white)$

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    $\begingroup$ 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

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