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Suppose there is a box with $N$ balls, $k$ white and $N-k$ black.

  1. If after choosing a ball it gets returned to the box, then: $$p(\text{white})=\frac{k}{N},\space\space\space p(\text{black}) = \frac{N-k}{N} = 1-p(\text{white})$$ — probabilities of choosing one white (black) ball, at each moment, regardless of how many balls will be selected.
  2. What if after choosing a ball, it will be kept outside the box, and the process of selecting balls will be continued, untill all balls are selected? I am aware that this is basic, course question. I am hoping for an answer or reference, to get a level of certainty.

  3. A third method of selecting balls is possible: suppose there is a simple device – a metal "matrix" with $N$ holes, in the shape of the box. The matrix is put on top of the box, which is then turned upside down. Also, we somehow ensure, that each hole will pass through exactly one ball. Then:

    • Process is similar to case 2., because balls are not returned to box after choosing.
    • Probability that given cell in the matrix will pass through white (black) ball is the same as in case 1., the basic case.
    • This means, that it is not the act of not returning ball that influences probabilities in case 2. – it is the act of receiving information about ball chosen, and thus missing (in case 2.) in the box.
    • Without recording the information, probabilities of case 2. are the same as in case 1.?
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up vote 2 down vote accepted

Suppose that we do not return balls to the box. The plain probability that the $i$-th ball selected is white is $\frac{k}{N}$, exactly like in the case of returning the ball to the box.

One way of seeing this is to number the balls, white from $1$ to $k$, black from $k+1$ to $N$. Imagine that we draw the balls, one at a time, until all the balls are gone. All permutations of the labels are equally likely, and the fraction of these permutations for which the $i$-th ball drawn is white is $\dfrac{k}{N}$. (It can take a while until this fact becomes "obvious"!)

But in drawing one at a time, there are other probabilities that we can compute, for example the probability that the $3$rd ball drawn is white given that the first two were white. This conditional probability is not $\dfrac{k}{N}$, it is $\dfrac{k-2}{N-2}$ (except in trivial cases, like $k=1$).

Suppose again that we draw the balls, one at a time, until they are all gone. The conditional probability that the $3$rd ball drawn is white, given that the last two balls drawn (of the $N$) are white is also $\dfrac{k-2}{N-2}$. So it is not the temporal order of the drawing that matters in evaluating the conditional probability.

When we calculate a conditional probability, it is not the act of receiving information that matters, it is the act of using the information to calculate a conditional probability, that is, to restrict the sample space.

When one does replacement, conditional and unconditional probabilities are the same, since the sample space is unchanged.

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I think I understand what you mean by the "fraction-argument". Uniform action (each ball can be selected equally likely) applied to given partition (1 to k; k+1 to N) does not influence the partition - because it is uniform. Correct? Then, I still have doubts about utilising the information about already selected balls – e.g. if white ball has been drawed, then we know that new probability p(white) is $\frac{k-1}{N-1}$, etc. But this will be only a short "fluctuation" to the probability $\frac{k}{N}$? – Calikin May 22 '12 at 18:48
Great answers, thanks. Are there maybe some sources about the topic? – Calikin May 22 '12 at 21:06
Andre: probability of drawing white, and then black ball, is of course $p(w) \cdot p(b)$, in both with/without returning cases? – Calikin May 23 '12 at 16:48
So in (ii) case, there are two events $p(w)$ –probability of drawing white ball, and $p(b|w)$ – probability of drawing black ball if one white has been already drawed (without replacement). I think I am beginning to understand the rule. The condition ("|w" in the conditional probaility) is the information that we are using to limit the sample space. If we will draw two balls at a time, then (ii) will equal to (i)? – Calikin May 23 '12 at 17:26
PS. I was hoping for reasoning similar to the first comment's: "fraction 1..k; k+1..N"..., so that drawing 2 balls white & red would be: $2\frac{k}{N}\frac{N-k}{N}$, and drawing white ball with left hand, black ball in right hand would be $\frac{k}{N}\frac{N-k}{N}$ because left and right hand is not really order. PS. I now see that your comment is exactly the same result as mine previous. – Calikin May 23 '12 at 17:51

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