Choosing 10 balls out of a box of 90 balls

Question

A box contains $ 30$ red balls, $30$ white balls and $30$ blue balls. If $10$ balls are selected at random without replacement, what is the probability that at least one color will be missing from the selection?

The answer is:

Solution: Let $A_1, A_2$ and $A_3$ be the events that there is no red, no white and no blue balls, respectively. Then by the inclusion-exclusion principle,

$P(A_1\cup A_2\cup A_3) = \sum_{i}P(A_i)-\sum_{i<j}P(A_i\cap A_j)+ P(A_1\cap A_2\cap A_3)$.

Clearly, $P(A_1\cap A_2\cap A_3)=0$ and $P(A_i\cap A_j)={\binom{30}{10}\over{\binom{90}{10}}}$ for $i\ne j$. Finally,

$P(A_i)= {{\sum\limits_{k=1}^{10}\binom{30}{k}\binom{30}{10-k}}\over{\binom{90}{10}}}={{\sum\limits_{k=0}^{10}\binom{30}{k}\binom{30}{10-k}-\binom{30}{10}}\over{\binom{90}{10}}}={{\binom{60}{10}-\binom{30}{10}}\over{\binom{90}{10}}}$

Why isn't it just $\binom{60}{10}\over\binom{90}{10}$?

Could you explain how do you get the result that you think is true? I will be easier to discuss about your problem (whatever the problem is in fact...). — user37238, Apr 16 '15 at 10:22
My question was : Explain us why do you think this is not the correct answer and why yours should be the correct one. — user37238, Apr 16 '15 at 10:44
because now you need to choose 10 out if 60 balls. why to subtract 30 choose 10 — Dkova, Apr 16 '15 at 11:40
@Dkova I agree with you: I think $P(A_i)=\binom{60}{10}\bigg/\binom{90}{10}$. — Mick A, Apr 16 '15 at 12:04

score 2 · Accepted Answer · answered Apr 16 '15 at 18:31

You are correct. The numerator

$$\sum_{k=1}^{10}\binom{30}k\binom{30}{10-k}\tag{1}$$

is, for instance, the number of ways to choose $30$ balls so that no ball is red, and at least one ball is white. This quantity is not useful in this problem. I could see someone looking at

$$\sum_{k=1}^{9}\binom{30}k\binom{30}{10-k}=\binom{60}{10}-2\binom{30}{10}\;,$$

which is, for instance, the number of ways to choose $30$ red and white balls so that you have at least one of each color: three times this quantity is the number of ways to pick $30$ balls and end up with exactly two colors represented amongst the $30$. One could then add $3\binom{30}{10}$, the number of ways to choose a monochromatic set, to get the number of sets containing balls of at most two colors. But $(1)$ makes no sense here that I can see.

Choosing 10 balls out of a box of 90 balls

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