could you please help me?

I know the correct solutions which are different from mine, and I absolutely agree that they are correct. However, I cannot find out what's wrong with my solutions that typed below.


Five balls are randomly chosen, without replacement, from an urn that contains 5 red, 6 white, and 7 blue balls. Find the probability that at least one ball of each color is chosen.

Sample space consists of $\binom{18}5$ outcomes. To meet the requirement I need to choose red ball by $\binom51$ ways, white - $\binom61$ and blue - $\binom71$. At the moment we have 3 out of 5 balls. The final balls could be drawn by $\binom{18-3}2$. Thus, the answer is

$$\binom51 \binom61 \binom71 \binom{15}2 / \binom{18}5$$

The most terrible thing about the above solution is that the probability exceeds 1.


Balls are randomly removed from an urn initially containing 20 red and 10 blue balls. What is the probability that all of the red balls are removed before all of the blue ones have been removed?

Imagine 30-element vector that could be filled with either letter "r" or "b". Hence, sample space consists of $2^{30}$ outcomes.

Obviously, the last element in this vector must be "b". That leaves 29 positions to be occupied, i.e. $2^{29}$ outcomes.

The probability is $2^{29}/2^{30} = \frac12$

Could you help me to find flaw(s) in my reasoning?


For 1: I think it fails because you're using different partitions for your balls. First, you say you're partitioning your balls into the red (size: 5), white (size: 6) and blue (size: 7) balls, and using binomial coefficients to count the different possibilities for each set. But then you're partitioning the balls into the ones you've already chosen (size: 3) and the ones you haven't (size: 15). I think choosing different partitions is problematic in itself with these type of counting and multiplication calculations, and partitioning based on "having already chosen" balls also leads to problems.

For 2: note that not all 30-element vectors are valid outcomes. For example, if the first 20 elements are "r", then there is only one choice for the remaining 10 elements.

  • $\begingroup$ Now I see my mistakes. Thank you, Mark, very much. 1) I've found the way to count it right: there are six cases of how many balls of each color may be included into sample: for example, 3 red & 1 white & 1 blue, i.e. (3,1,1) for short. It gives $\binom{5}{3}\binom{6}{1}\binom{7}{1}$ outcomes. And so on. Then, I added them, divided by total # of outcomes and got the correct answer. $\endgroup$ – Yal dc Nov 20 '13 at 19:25
  • $\begingroup$ Glad I could help! $\endgroup$ – Mark Nov 20 '13 at 20:48

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