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We distribute 10 indistinguishable balls to 5 girls.

All the distributions have equal probability

Let X be the number of girls who get at least 1 ball

I need to find $Pr(X=3)$ and $E[X]$

But there are a few things I did not understand.

  1. Why do I need the info that all the distributions have equal probability? What would have happen without it?

  2. I can't find the right module to use here, I thought it's related to binomial distribution but seems like I can't apply it properly, I probably don't understand it enough -> how & when to use it and if here is the right case?

Thanks in advance

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The One
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  • The fact that all distributions have the same probability means that the probability of any one event occurring is $1/\text{Total number of distributions}$. In other cases, this formula would be incorrect. Basically, the statement is saying what process is used to distribute the balls. – Michael Burr Apr 29 '16 at 10:39
  • @MichaelBurr so each girl has $0.5$ chance to get a ball and $1-(0.5)^{10}=0.999$ to get at least 1 ball, but the fact that a given girl got at least 1 ball will affect my other calculations so I don't know how to find $X=3$ – The One Apr 29 '16 at 10:50
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    The answer to "What would have happened without it?" is: The question wouldn't have been well-defined, since there's no meaning to talking about probabilities and expectation values without assuming some probability distribution. Sometimes problems don't specify one because the author (often wrongly) believes that a certain distribution is clearly implied. In the present case, the most natural assumption if no distribution had been specified might have been that each ball is given to any one of the girls with equal probability $\frac15$. – joriki Apr 29 '16 at 11:15
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    In that case, the indistinguishable distributions that are taken as equiprobable here would not have been equiprobable. In this sense, this specification makes a difference. – joriki Apr 29 '16 at 11:16

2 Answers2

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Hints:

  1. What is the total number of distributions? Let $x_1$, $x_2$, $x_3$, $x_4$, and $x_5$ be the number of balls that each girl receives (I assume that I can tell the girls apart). We know that $x_1+x_2+x_3+x_4+x_5=10$ and they are all nonnegative integers. Therefore, the number of values for $x_1,\cdots,x_5$ is the number of sums of $5$ nonnegative integers equalling $10$. This is a standard problem (related to the "stars and bars" problem (not the flag)), one solutions can be found here. In this case, $n=10$ and $k=5$, and the standard count is $\binom{n+k-1}{k-1}=\binom{14}{4}=1001$.

  2. To compute $Pr(X=3)$, first consider what $X=3$ means; there are exactly three girls who have balls and two which do not. In the sum above, exactly two of $x_1,\cdots,x_5$ are zero and exactly three are nonzero. The first question is which three are nonzero? There are $\binom{5}{3}=10$ ways to select these three girls.

  3. Now, you've selected which three girls will receive at least one ball. Assume that these three are $(i,j,k)$. Therefore $x_i$, $x_j$, and $x_k$ are all positive and $x_i+x_j+x_k=10$. This is another "stars and bars" problem, but the number of ways that this can occur is $\binom{10-1}{3-1}=\binom{9}{2}=36$.

Therefore, the number of ways to have exactly three girls receive at least one ball is $10\cdot 36=360$. Since there were $1,001$ ways to distribute the balls, the probability $Pr(X=3)=\frac{360}{1001}$. To compute the expectation, generalize the ideas in this approach.

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Michael Burr
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  • Thanks does each time I get a problem like this I need to calculate it by hand? since I know there are some modules that apply to certain problems where you know the formula of P(X) and the Var and E[X]. Also for the expectation do I need to find X=1,2,4,5 now and work by definition? or there is a faster way than this? – The One Apr 29 '16 at 11:19
  • Usually a problem is not a "standard type" as you describe, but it can be broken down into standard problems (like what is done above). So, I guess the answer to your first question is yes and no? Yes, you will need to compute the probabilities $Pr(X=i)$ for $i=1,2,4,5$ to compute the expectation. If you understand the case where $Pr(X=3)$, then you can use the same ideas and construction to make quick work of the other cases. – Michael Burr Apr 29 '16 at 11:22
  • I have a question: If we let Y to be the number of girls who got 0 balls can I say that the Var(Y) = Var(X)? because only 1 girl that gets at least 1 ball is the same case as exactly 4 girls that do not get any so I just felt like I could calculate Var(X) instead of now doing all over again with Y, is it right? – The One Apr 29 '16 at 11:53
  • If you've already computed everything for $X$, then you can easily compute $Y$ because $Y=5-X$, so the probabilities are the same, but the values of the random variables are different. – Michael Burr Apr 29 '16 at 12:05
  • I've found that $Var(X) = 0.705$ and $E[X]=3.571$ so I said that $Var(Y)=0.705$ since they are the same by formula but it seems like it is wrong the only closest answer is 0.71 and I don't understand why I'm off by 0.005 or its ok to be wrong in such a small margin? – The One Apr 29 '16 at 13:00
  • Do they ask for the answer to be rounded? – Michael Burr Apr 29 '16 at 13:04
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Hint.- 10 balls to five girls, consider:

10-0-0-0-0

9-1-0-0-0

8-2-0-0-0

8-1-1-0-0

... ...

cgiovanardi
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