Question :
In a box there are 50 red and 200 white balls. Somebody removes 10 randomly selected balls.What is the probability that no red ball is among these removed balls?Do we need any assumption to solve this problem?
Expected Answer :
using binomial distribution: $0.05631351$
using sampling without replacement : $0.05209363$
My approach : To select no red balls, they should all be white that is: $$C(200,10)/C(250,10) = 0.1025135$$ Which does not align with the answer
What kind of an assumption do I need to solve this problem?
How are binomial and sampling without replacement different?
Most importantly how to solve this problem?

  • 2
    $\begingroup$ What do you mean by "binomial distribution" as a method here? If we were sampling with replacement or if you just make the simplifying assumption that each draw is independently white with probability $.8$ then the answer would be $.8^{10}=0.107374$ which is in line with your method. $\endgroup$ – lulu Feb 15 '20 at 12:14
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    $\begingroup$ I think the only "assumption" that matters is whether you are drawing with or without replacement. As the problem is phrased I'd have assumed it was "without replacement" . $\endgroup$ – lulu Feb 15 '20 at 12:17
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    $\begingroup$ I don't understand either of the two proposed answers. Your answer looks solid, trusting that "without replacement" was intended. $\endgroup$ – lulu Feb 15 '20 at 12:20
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    $\begingroup$ Your method and answer look good for the without replacement case. And with replacement it's simply $(\frac{200}{250})^{10} \approx 0.10737$, quite close to the other situation. $\endgroup$ – Deepak Feb 15 '20 at 12:23
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    $\begingroup$ The exact solution is Hypergeometric distribution. $\endgroup$ – Lee David Chung Lin Feb 15 '20 at 14:10

\begin{align} &\frac{ (\text{choose 0 from 50 red}) \times (\text{choose 10 from 200 white})}{ \text{choose 10 from totally 250}} \\ &= \frac{C(50,0)\times C(200,10)}{C(250,10)} \qquad \text{or in U.S. notation} \quad\frac{ {50 \choose 0}\times {10 \choose 200}}{ 10 \choose 250} \\ &= \frac{1 \times 22451004309013280}{219005316087032475}\\ &\approx 0.10251351296007479218513634 \end{align} This is the extreme case (zero red balls) of the discrete Hypergeometric distribution, one of the common distributions that sometimes gets forgotten.

For example, if the question asks for the probability of $3$ red balls among those $10$, the answer is

$$\frac{C(50,3)\cdot C(200, 7)}{C(250,10)}$$

In Binomial distribution, the trials (draws) are independent to each other and the "success" probability $p$ (here the fraction of red ball among all balls) stays constant. When handling finite population (number of balls) and without replacement, the Hypergeometric is exact, and we know that the $p$, the fraction of red balls, keeps changing after each draw (removal).

When the number of two types of items (red balls, white balls) are both large "enough" compared with the number of removal so that $p$ stays roughly constant, then the Binomial distribution is a good approximation (mathematically there's a limit theorem) to the Hypergeometric distribution.

Here, for example, the total number of red balls is $50$ (which is smaller then the $200$ for white), and $50$ is considerably larger $10$ so the Binomial approximation is sort of not too bad. The fraction of red balls remains roughly $p\approx 50/250 = 0.2$. One can check that the "worst" case is for $p$ to drop to $(50-9)/(250-9) \approx 0.17$


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