# Is this answer to this probability question correct?

A class of $400$.
• $220$ women.
• $180$ men.
Divided into study groups of 10 students each.
Q. What is the probability that a certain group will have exactly $4$ men?

$$P(x) = \frac{n!}{x! \cdot (n-x)!} \cdot p^x \cdot (1-p)^{(n-x)}$$

$n=4 \qquad x=4 \qquad p=0.45$

$P(x) = \frac{10!}{4! \cdot (10-4)!} \cdot 0.45^4 \cdot (1-0.45)^{(10-4)}$

$\qquad \,\, = 23.84\%$

This is from the Binomial Distributions topic of a Udemy course I'm taking. That's the given answer. This seems like how to find the probability of a certain group having four men if each person was chosen from $10$ different groups where probability of choosing a man is $0.45$.

So, is that answer correct? What am I missing?

• This is random variable $X$ follows the hypergeometric distribution (without replacement) rather than the binomial distribution: $$P(X=4)=\frac{\binom{180}{4}\cdot \binom{220}{10-4}}{\binom{400}{10}}=24.1\%$$ – callculus Feb 17 '18 at 16:12
• I appreciate that you recognized this flaw in the video. Respect. I would you give 10 upvotes more if I could. – callculus Feb 17 '18 at 19:19

Lets make a small example: You have a group of 2 men and 3 woman. Now you want divide into a study group of 3 students with 2 women ($Y$). The ways to do that is

$\texttt{wwm, wmw, mww}$

Each group has the same probability. Let us focus on the first group. The probability to choose a woman first is $\frac{3}{2+3}=\frac{3}{5}$

The probability to choose woman again is $\frac{3-1}{2+3-1}=\frac{2}{4}=\frac{1}{2}$

Here you can see that you have to use a distribution which regard, that the woman are not replaced.

The probability to choose a man is $\frac{2}{2+3-2}=\frac{2}{3}$

Thus the probability to choose $\texttt{wwm}$ is $\frac{3}{5}\cdot \frac{1}{2}\cdot \frac{2}{3}=\frac1{5}=20\%$.

Since we have three ways to choose $2$ woman, the probability to choose $2$ woman is $P(Y=2)=60\%$. This is equal to $$P(Y=2)=\frac{\binom{3}{2}\cdot \binom{2}{1}}{\binom{5}{3}}$$

It seems that the given example is supposed to show that the binomial distribution can be used as an approximation for the hyper-geometric one if

• the number of selected items is small in comparision with the size of the population one selects from
• the probability $p$ of a "success" (here man in group) is relatively stable if only a few items (men) are selected - this is so if $p$ is not too far from $0.5$

As callculus indicated above, the correct solution using the hyper-geometric distribution is $$\frac{\binom{180}{4}\cdot \binom{220}{10-4}}{\binom{400}{10}} \approx 0.241$$

The approximation using the binomial distribution with $p = \frac{180}{400} = 0.45$ is $$\binom{10}{4} \cdot 0.45^4 \cdot 0.55^6 \approx 0.238$$

• But in this part of the video the approximation is not mentioned. So this is a clear mistake. – callculus Feb 17 '18 at 18:50
• Sad to hear. Another case where "educational math" produces erroneous pseudo-math and confuses those learners who really want to understand. – trancelocation Feb 17 '18 at 19:03
• That´s really true. And also: Academic titles sometimes means nothing: He is an Associate Dean of Executive MBA Programs at UCLA Anderson School of Management. Here is the link to the video. – callculus Feb 17 '18 at 19:15
• The video is not listed any more on his channel. I good decision from George. But it can be still found at the link – callculus Jul 15 at 19:11