Binomial coefficient real life example.

"At a university, $15$ juniors and $20$ seniors volunteer to serve as a special committee that requires $8$ members. A lottery is used to select the committee from among the volunteers. Suppose the chosen students consists of six juniors and two seniors.

(a) For a test of homogeneity, what are the expected counts?

This question I understand.

(b) If the selection had been random, what is the probability of the committee having exactly two seniors?

My answer was that the probability is binomial, with Binom$(k=2, n=8, p=0.57)$, but this is apparently wrong. Instead the correct answer is: $$\frac{\binom{20}{2}\binom{15}{6}}{\binom{35}{8}}$$.

Can anyone explain the difference between this and standard binomial distribution?

• I suppose that you made a type and that there are $15$ junior volunteers, not $13$. Am I right? – José Carlos Santos Jun 17 '18 at 7:12
• It's not a straight binomial since the trials are not independent. Knowing that the first choice was a senior changes the probability that the second was a senior. – lulu Jun 17 '18 at 7:12
• Ahh. Thanks! I understand now :-) – Mathe Jun 17 '18 at 7:20
• This is the hypergeometric distribution – Lord Shark the Unknown Jun 17 '18 at 7:21
• To be sure of useful answers, you should state null and alternative hypotheses for your 'test of homogeneity'. You don't have quite large enough expected counts for the chi-squared goodness-of-fit statistic to truly have a chi-squared distribution. // Using the hypergeometric distribution, it seems you are aiming at "Fisher's exact test' which you can google. – BruceET Jun 17 '18 at 8:04

The issue is that the trials are not independent from each other. Having chosen a junior first, for example, changes the probability of now choosing a senior.

You must look at how many ways are there to choose two seniors (which is exactly $\binom{20}{2}$) and how many ways are there to choose six juniors (which is exactly $\binom{15}{6}$) and multiply them. This gives you the overall number of valid arrangements.

To get the probability, simply divide by the number of overall arrangements possible (which is $\binom{35}{8}$).

• To get a p-value you also need to include the probabilities of more extreme cases. – BruceET Jun 17 '18 at 8:47

In case it is helpful, here is Minitab output for testing two proportions. It attempts a normal test (with a warning about sample sizes being too small) and does Fisher's exact test.

Test and CI for Two Proportions

Sample  X   N  Sample p
1       6  15  0.400000
2       2  20  0.100000

Difference = p (1) - p (2)
Estimate for difference:  0.3
95% CI for difference:  (0.0193759, 0.580624)
Test for difference = 0 (vs ≠ 0):  Z = 2.10  P-Value = 0.036

* NOTE * The normal approximation may be inaccurate for small samples.

Fisher’s exact test: P-Value = 0.051