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  1. Can you help me with probability problem please?

There are 3 boxes. In the first box are 2 black balls and 1 white ball. In the second box are 2 white balls and three black balls. The third box is empty. 2 balls are randomly chosen from the first box and put in the third. 2 balls are randomly chosen from the second box and put in the third. So there are 4 unknown balls in third box.

What is the probability of choosing 1 white ball from third box?

What is the probability of putting two black balls from first box into the third if one black ball is chosen from third box? if two black balls are chosen from third box?

  1. Is there a common approach to solving these problems?
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For your first problem, we need to get a handle on what the third box can look like. There are $\binom{3}{2}=3$ to choose the two balls from the first box and there are $\binom{5}{2}=10$ to choose the two balls from the second box. Thus there are $\binom{3}{2}\binom{5}{2}=30$ ways for the contents of the third box to be selected.

The third box can only look like one of the four following possibilities: $BBBB$, $WBBB$, $WWBB$, or $WWWB$. Notice we can't have 4 white balls since there are only a total of 3 white balls to choose from. So our $30$ configurations we counted previously must be split up between these 4 descriptions. So how many of each kind are there?

$BBBB$: The only way this can happen is if we select two black from the first box and two black from the second box. So there are $\binom{2}{2}\binom{3}{2}=3$ ways for this to happen.

$WBBB$: 1.) Select one white/one black from the first box and two black from the second box for a total of $2\cdot \binom{3}{2}=6$ ways. or

2.) Two black from the first and one white/one black from the second box for a total of $1\cdot (3\cdot 2)=6 ways.

Thus there are a total of $6+6=12$ ways for this to happen.

$WWBB$: 1.) Select one white/one black from the first box and one white/one black from the second box for a total of $2\cdot (3\cdot 2)=12$ ways.

2.) Two black from the first and two white from the second box for a total of $1\cdot\binom{2}{2}=1$ ways.

Thus there are a total of $12+1=13$ ways for this to happen.

$WWWB$: Select one white/one black from the first box and two white from the second box for a total of $2\cdot \binom{2}{2}=2$ ways.

From the above work we know that $P(BBBB)=\frac{3}{30}$, $P(WBBB)=\frac{12}{30}$, $P(WWBB)=\frac{13}{30}$, and $P(WWWB)=\frac{2}{30}$

So the probability we draw a white ball from the third box can be calculated by :

$$ \begin{align*} P(W)&=P(W\,\cap\,WBBB)+P(W\,\cap\,WWBB)+P(W\,\cap\,WWWB)\\ &=P(W\,\vert\,WBBB)\cdot P(WBBB)+P(W\,\vert\,WWBB)\cdot P(WWBB)+P(W\,\vert\,WWWB)\cdot P(WWWB)\\ &=\frac{1}{4}\cdot\frac{12}{30}+\frac{2}{4}\cdot\frac{13}{30}+\frac{3}{4}\cdot\frac{2}{30}\\ &=\frac{12+26+6}{120}\\ &=\frac{11}{30}. \end{align*} $$

For your second question, the analysis leading up to the answer for the answer to the first question should be sufficient to allow you to figure that out.

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