Assume there is a drawer with $4$ balls of different colors. You draw lots $3$ times and put the ball back after each draw.

What is the probability that the same ball will be drawn exactly $1$ time?

My 'guess': I would use the binomial formula:

$$ P_{1} = \binom{3}{1} \cdot \left(\frac{1}{4}\right)^{\mkern -4mu 1} \cdot \left(1 - \frac{1}{4}\right)^{\mkern -4mu 3 - 1} = 3 \cdot \frac{1}{4} \cdot \frac{9}{16} = \frac{27}{64}. $$

Is this correct?


It seems that I got the correct answer. Now to a related question:

Why is the probability of drawing the same ball exactly $0$ times the same the probability of drawing it exactly $1$ time? This is now intuitively clear to me.

  • $\begingroup$ You are absolutely right. $\endgroup$ – Satish Ramanathan Jun 6 '14 at 13:43
  • $\begingroup$ This is the probability that a specific ball (the blue one) is drawn exactly once. If that is what you intended to compute, the answer is correct. $\endgroup$ – André Nicolas Jun 6 '14 at 13:43
  • $\begingroup$ $\frac{27}{64} \ne 0.14$ $\endgroup$ – Alex Jun 6 '14 at 14:05

From your follow up question, I consider that you mean a specific ball, that you want to draw exactly once (original question) or not of all (follow up.)

Let us first generalize the problem: We have a urn with $n$ balls. One of the ball if marked. We draw $n-1$ at put the balls we draw back into the urn.


  1. What is the probability to draw the marked ball exactly once?
  2. What is the probability to not draw the marked ball at all?


  1. $ p_1 = {n-1 \choose 1} \cdot \frac{1}{n} \cdot (\frac{n-1}{n})^{n-2}$
  2. $ p_2 = (\frac{n-1}{n})^{n-1}$

As you noticed the two solutions coincide. This is due to following equality: $$ {n-1 \choose 1} \cdot \frac{1}{n} = \frac{(n-1)!}{(n-2)! 1!} \cdot \frac{1}{n} = (n-1) \cdot \frac{1}{n} = \frac{n-1}{n} $$

  • $\begingroup$ Great. Thank you very much! $\endgroup$ – Svend Tveskæg Jun 6 '14 at 14:56
  • $\begingroup$ where's the inequality? $\endgroup$ – Alex Jun 6 '14 at 15:02
  • $\begingroup$ @Alex : Your right! My fault, it should be equality instead of inequality. I changed my answer accordingly. Thanks. $\endgroup$ – crixstox Jun 16 '14 at 9:37

This is called sampling with replacement. Assuming I correctly understand your question ('same ball exactly 1 time') what you need is the probability of draws like $ABC, ABD, ACD, BCD$ - there are totally 4 unique combinations and each combination has $3!$ representations (i.e. ABC, BCA...CBA), hence there are $3!4 = 4!$ 'positive outcomes' out of $4^3$ total outcomes. Hence your probability is $\frac{4!}{4^3}$.

  • $\begingroup$ Just another reasoning with the same result: For the first draw there are no restriction, for the second one you can should any off the three remaining balls (sucess probability $\frac{3}{4}$) and in the last drawing round any off the two remaining balls. Overall you have the probability: $p = \frac{3}{4} \cdot \frac{2}{4}$, which coincides with the answer from @Alex. $\endgroup$ – crixstox Jun 6 '14 at 14:03
  • $\begingroup$ no, the result is different. $\frac{3}{8} \ne 27/64 \ne 0.14$. The question is quite vague though. $\endgroup$ – Alex Jun 6 '14 at 14:05

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