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Suppose we have a urn with $24$ red balls and $8$ black balls. We randomly draw $n$ balls from the urn without putting them back. How big must $n$ be such that the probability of drawing at least $2$ red balls is $0.9$?

The probability is the same as $1$ minus the probability of drawing at most $1$ red ball, which is perhaps easier to find. But how do I find it?

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  • $\begingroup$ "But how do I find it?" As the sum of the probabilities of getting 1 red ball and no red balls. $\endgroup$ – leonbloy Nov 6 '14 at 16:01
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Not so elegant, but I think this is the fastest way to answer this question (that is, obtain a minimal value of $n$ such that the probability of having at least $2$ balls is more than or equal to $0.9$)

By constraints of the question we have $0 \leq n \leq 32$. Furthermore there are $8$ blue balls, and $24$ red balls.

So it is obvious that P( at least 2 are red ) increases as $n$ increases. In fact, by the Pigeonhole Principle, for $ 10 \leq n \leq 32 $ the probability of having at least $2$ red balls is $1$.

For $n=0$ or $n=1$, it is impossible to have at least $2$ red balls. The probability is $0$.

For $n=2$, there is only the case $2$ out of $2$ of the balls being red to consider. So we get:

$\dfrac{(24 \cdot 23)}{(32 \cdot 31)} \approx 0.55645$

For $n = 3$, we need to consider $2$ out of $3$, as well as $3$ out of $3$ of the balls being red. So we get:

$\dfrac{(8)(24 \cdot 23) {3 \choose 2} }{( 32 \cdot 31 \cdot 30 )} + \dfrac{(24 \cdot 23 \cdot 22)}{32 \cdot 31 \cdot 30} \approx 0.85323$

Now for $n=4$, we need to consider $2$ out of $4$, $3$ out of $4$, as well as $2$ out of $4$ of the balls being red. So we get:

$\dfrac{(8 \cdot 7 )(24 \cdot 23) {4 \choose 2} }{( 32 \cdot 31 \cdot 30 \cdot 29 )} + \dfrac{(8)(24 \cdot 23 \cdot 22) {4 \choose 3}}{( 32 \cdot 31 \cdot 30 \cdot 29)} + \dfrac{(24 \cdot 23 \cdot 22 \cdot 21)}{( 32 \cdot 31 \cdot 30 \cdot 29)} \approx 0.96068 $

Hence the answer is $n=4$.

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  • $\begingroup$ Thanks a lot!! But how did you get the term $$\dfrac{(8)(24 \cdot 23) {3 \choose 2} }{( 32 \cdot 31 \cdot 30 )}$$? I know it's supposed to be getting 2 out of 3 red balls, but I dont understand it. Thank you. $\endgroup$ – user190397 Nov 7 '14 at 7:58
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There are $\binom{32}{n}$ possibilities when it comes to drawing $n$ balls out of $32$.

Under the condition that exactly $r$ of these balls are red there are $\binom{24}{r}\binom{8}{n-r}$ possibilities.

If $R_{n}$ denotes the number of red balls that are drawn here then $$P\left(R_{n}=r\right)=\binom{24}{r}\binom{8}{n-r}\binom{32}{n}^{-1}$$

To be found is the minimal $n\geq2$ with $$P\left(R_{n}\geq2\right)\geq0.9$$ or equivalently with: $$P\left(R_{n}=0\right)+P\left(R_{n}=1\right)\leq0.1$$

That comes to finding the minimal $n\geq2$ with $$\binom{24}{0}\binom{8}{n}\binom{32}{n}^{-1}+\binom{24}{1}\binom{8}{n-1}\binom{32}{n}^{-1}\leq0.1$$

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You have

$$P(\text{at least 2 reds}) = 1 - P(\text{no reds}) - P(\text{exactly 1 red})$$

so

$$P(\text {at least 2 reds}) = 1 - {{{8\choose n}}+ {8\choose n-1}\cdot 24 \over {32\choose n}} $$ provided $n \ge 2$. Now find $n$ so that the subtracted quantity is less than $.1$.

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  • $\begingroup$ Thanks. I don't understand how you got the term $${{{8\choose n}}+ {8\choose n-1}\cdot 16 \over {24\choose n}}$$ Could you kindly elaborate on what it means? $\endgroup$ – user190397 Nov 6 '14 at 16:12
  • $\begingroup$ There are not $16$ but $24$ red balls. $\endgroup$ – drhab Nov 6 '14 at 17:02

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