A box contains $20$ balls all of different colors including the red color. If we select $10$ balls randomly without replacement, what is the probability that the red ball will be among these $10$ balls?

What I think is that: If we let $X$ to be the number of balls we select until we get the red ball, then $X$ will be a random variable with range $ 1,2,3, \ldots ,20 $, and the probability of getting the red ball will be $1/20$, so our probability will be $$\left(\frac{1}{20} \right)^{10} ,$$ is that right?! I'm not sure about the distribution of $X$?

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    $\begingroup$ Your argument is wrong. But how do we check that? Ok, instead of picking just 10 balls, suppose I pick all the 20 balls. Without any calculations, what do you think is the probability the red ball is among the 20 balls? Now, what does your argument tell you? $\endgroup$ – Srivatsan Sep 19 '11 at 2:10
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    $\begingroup$ If you select all the 20 balls then the probability is 1. So my argument is false! $\endgroup$ – Kelly Sep 19 '11 at 2:15
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    $\begingroup$ Not just that. As you pick more balls, the probability should clearly increase. Your answer $$\left(\frac{1}{20} \right)^{\text{number of balls picked}}$$ is decreasing as I pick more balls and so it must be wrong. (If you change the answer to $1$ minus that value, then it is still wrong, but it is slightly better.) $\endgroup$ – Srivatsan Sep 19 '11 at 2:19

Your calculation represents the probability of getting the red ball ten times in a row with replacement: this is a totally different problem. It should be easy to see that it is false by how small it is: the probability of getting the red in just the first pick is one out of twenty, so your chances of getting the red ball in the process of taking ten balls out should be even greater.

Here's how the reasoning with $X$ could have gone: the chance the $n$-th ball came up red is the probability the other $n-1$ before it came up nonred and then you got the red ball out of the remaining ones (the latter colored red in the calculations below). The probability the first pick was nonred is $19/20$, the probability the second pick was nonred is $18/20$, and so on, so we get the pattern

  • $P(X=1)=\color{Red}{\frac{1}{20}}=1/20$
  • $P(X=2)=\frac{19}{20}\times\color{Red}{\frac{1}{19}}=1/20$
  • $P(X=3)=\frac{19}{20}\frac{18}{19}\times\color{Red}{\frac{1}{18}}=1/20$
  • $P(X=4)=\frac{19}{20}\frac{18}{19}\frac{17}{18}\times\color{Red}{\frac{1}{17}}=1/20$
  • $\qquad\qquad\cdots\cdots\cdots\cdots\cdots\cdots$
  • $P(X=10)=\frac{19}{20}\frac{18}{19}\cdots\frac{11}{12}\times\color{Red}{\frac{1}{11}}=1/20$

Hence $P(X=1)+P(X=2)+\cdots+P(X=10)=\frac{1}{20}+\cdots\frac{1}{20}=10(\frac{1}{20})=1/2.$

Alternatively, we could have reasoned with symmetry as follows: every way of picking twenty balls out of twenty balls and keeping track of what order they come out will be like putting the balls in some order. The probability a specific ball will end up in a specific position is equal to the same probability for any other specific ball, say $p$. Since at least one ball must be in the position we have that the sum of each ball's probability being there equals one, or $20p=1$, hence $p=1/20$. Sum this over the first ten positions for the red ball and you get $10(1/20)=1/2$ as the probability the red ball is in the first ten positions. Since it makes no difference whether we actually take out the last ten balls or keep them in the box, this must be our desired value.

Two quicker ways could have been the following:

  1. Symmetry: For every way of taking 10 balls out of 20 and leaving the other 10 in the box, there is exactly one way of taking those other 10 out of the box and leaving the original 10 in, hence by symmetry the probability of choosing the red ball equals the probability of not choosing the red ball. Thus the probability is $1/2$.

  2. Counting: The number of ways of getting a red ball in a choice of 10 out of 20 is equal to the number of ways the other 9 non-red balls can be chosen out of the total 19 non-red balls: in other words, $19\choose9$. And the total number of ways of picking 10 balls out of 20 is $20\choose10$. So the probability of getting the red ball is $$\frac{{19\choose9}}{{20\choose10}}=\frac{19!}{9!10!} \frac{10!10!}{20!}=\frac{10}{20}=\frac{1}{2}.$$

Having these two techniques in your probability toolkit is important because problems that can be solved easily with them are ubiquitous in combinatorial situations. However it is also important that you don't split a probability problem into a large number of smaller cases (originally, we split ours up into 10 different cases) when doing so isn't actually necessary - this saves a lot of energy.

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  • $\begingroup$ I don't think symmetry works in this case because I have a second part of the problem which generalize this to any box with n different objects. Thanks! $\endgroup$ – Kelly Sep 19 '11 at 2:30
  • $\begingroup$ @Kelly: Symmetry does in fact work "in this case" - is my description of it in this particular case not clear enough, should I elaborate on it further? The idea may or may not be applicable to your general problem, but it is certainly applicable to this specific one. $\endgroup$ – anon Sep 19 '11 at 2:36
  • $\begingroup$ Does X: the number of balls we select until we get the red ball, has a well-known distribution? $\endgroup$ – Kelly Sep 19 '11 at 2:59
  • $\begingroup$ @Kelly: I've edited my answer to be more comprehensive; it establishes that $X$ is uniformly distributed first using direct computation and then with symmetry. Please tell me if something is unclear. $\endgroup$ – anon Sep 19 '11 at 3:54
  • $\begingroup$ @Kelly: If "number of balls we select until we get the red" includes the successful selection, we are dealing with a very special case of the negative hypergeometric distribution. In that distribution, we have a group of size $N$, and $g$ of the members of the group are good. We draw without replacement until we get $k$ good. The numbers $N$, $g$, $k$ are fixed. The random variable $X$ is the number of draws. In your case, $N=20$ and $g=k=1$. I don't know whether the case $k=1$ has a special name. $\endgroup$ – André Nicolas Sep 19 '11 at 6:25

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