I'm sitting with a task, in which I got the answer already. The task is the following:
"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?
2 Answers
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 $\binom{35}{8}$).
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
