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I have 3 white and 4 black balls and I put them into red, green, blue, and yellow bags.

What is the sample space and find the number of possible outcomes? Is the sample space just all combos in which the possible balls are put into each bag? How would I solve this?

What is the probability that exactly three balls end up in the blue bag? would this be 7 choose 3 times 4 choose 1 over the sample space?

What’s the probability that all white balls end up in the same bag? Is this asking the same thing as above but with just a different name?

Whats the probability that at least one of the bags empty? I can calculate the probability that no bag is empty. How would I attempt this.

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The basic formula

We have $n$ distinguishable bags and $k$ objects to be arranged. There are $$\begin{pmatrix} n+k-1\\ k\\ \end{pmatrix} .$$ possibilities. (see: http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29). I am going to use this fact many times below.

The sample space consists of all the possibilities that the $3$ white balls can be arranged in the $4$ distinguishable bags, times, all the possibilities that the $4$ black balls can be put into the same bags. That is, we have

$$\begin{pmatrix} 4+3-1\\ 3\\ \end{pmatrix} \begin{pmatrix} 4+4-1\\ 4\\ \end{pmatrix}=\begin{pmatrix} 6\\ 3\\ \end{pmatrix} \begin{pmatrix} 7\\ 4\\ \end{pmatrix}=700$$

equally likely arrangements.

The first question: There are $4$ possibilities to end up with $3$ balls in the blue bag:

(1) $3$ black balls in the blue bag. Then remain $1$ black ball and $3$ white balls to be arranged in the $3$ other bags. That is, we have $$\begin{pmatrix} 3+1-1\\ 1\\ \end{pmatrix} \begin{pmatrix} 3+3-1\\ 3\\ \end{pmatrix}=\begin{pmatrix} 3\\ 1\\ \end{pmatrix} \begin{pmatrix} 5\\ 3\\ \end{pmatrix}=30.$$ In what follows I will not detail the use of the basic formula.

(2) $2$ black balls and $1$ white ball in the blue bag: $$\begin{pmatrix} 4\\ 2\\ \end{pmatrix} \begin{pmatrix} 4\\ 2\\ \end{pmatrix}=36.$$

(3) $1$ black ball and 2 white balls in the blue bag: $$\begin{pmatrix} 6\\ 3\\ \end{pmatrix} \begin{pmatrix} 4\\ 1\\ \end{pmatrix}=80.$$

(4) $3$ white balls in the blue bag: $$\begin{pmatrix} 6\\ 3\\ \end{pmatrix} =35.$$

The bottom line is $181$ equally likely specific arrangements. The answer to this question is

$$\mathbb P(\text{There are exactly 3 balls in the blue bag})=\frac{181}{700}.$$

The second question: There are four possibilities that all the $3$ white balls end up in the same bag. Then the remaining black balls will have to be arranged in $4$ boxes. That is we have

$$4\begin{pmatrix} 7\\ 4\\ \end{pmatrix} =140.$$

The answer to this question is

$$\mathbb P(\text{All the 3 white balls are in the same bag})=\frac{140}{700}.$$

Third question The answer is apparently negative.

The fourth question:

(1) There are $4$ possibilities that exactly one bag is empty. Then we have $3$ white balls and $4$ black balls to be arranged in the remaining 3 bags in such a way that non of these boxes remain empty. The black balls have $3$ such arrangements and the white balls have only $1$ such an arrangement. The total is $12$.

(2) There are $6$ possibilities that exactly $2$ bags are empty and none of the remaining bags are supposed to be empty. For the black balls there are $3$ such possibilities and for the white balls there are only $2$ such possibilities. Then the number of all possibilities is $36$.

(3) If exactly $3$ bags are empty then we have $4$ possibilities but then all the balls will have to be in one bag. There are then $4$ possibilities only.

The answer to this question is

$$\mathbb P(\text{At least one bag is empty})=\frac{52}{700}.$$

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  • $\begingroup$ why do you use 6 choose 3 and then 7 choose 4 in the beginning. I understand 7 choose 4, but not 6, shouldnt it be 7? $\endgroup$ Feb 8 '15 at 17:44
  • $\begingroup$ OK I edit... nnn $\endgroup$
    – zoli
    Feb 8 '15 at 18:12
  • $\begingroup$ @Jack : I've edited. $\endgroup$
    – zoli
    Feb 8 '15 at 18:42

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