Question

An instructor who taught 2 sections of statistics last term, the first with 20 students and the scond with 35, decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects.

a). What is the probability that exactly 10 of these are from the second section?

b). What is the probability that at least 10 are from the second section?

c). What is the probability that at least 10 of them are from the same section?

If you could please elaborate on the steps that you took, that would be greatly appreciated!!

Answer 1

I know that a). can be easily solved using the form:
\[P(X=x)=\frac{ \left(\begin{matrix}M \\ x\end{matrix}\right)\left(\begin{matrix}N-M \\ n-x\end{matrix}\right) }{ \left(\begin{matrix}N \\ n\end{matrix}\right) }\]
Which I got to be 0.2392
For b). Must I do this for every x between (and including) 10-15 and add them?

I'm still confused on c).

Answer 2

For your solution to (a) you assume sampling without replacement. However if the instructor randomly ordered the 55 assignments, and then the first 15 were taken for grading, I think that this sample of 15 would follow a binomial distribution of the number from each section.

Answer 3

If the distribution is binomial, the required probability for (a) is:
\[\large P(10\ from\ second\ section)=\left(\begin{matrix}15 \\ 10\end{matrix}\right)\times(\frac{7}{11})^{10}\times(\frac{4}{11})^{5}=0.208\]

Answer 4

Actually, I didn't realize that the software gives you the answers (not the steps though) once the assignment has past its due date. The assignment was due a while ago, but I'm trying to review the problems I couldn't understand. The answer they gave for a). matched the one I put.

Answer 5

Wouldn't it be hypergeometric though? It could be defined as
h(x; 35, 15, 55). Then just plug in 10 for x for which we want to find the probability of, right?

Answer 6

Because we want to know the probability of the 10 "successes" in the subset 35 with 15 observations in a total population of 55.

Answer 7

If the instructor randomly ordered the 55 projects and then randomly selected projects for marking, then the hypergeometric distribution applies. This would indeed be sampling without replacement.

Answer 8

How do you mean `without replacement` ?

Answer 9

After the instructor takes the first project and marks it, the marked project is not replaced. There are now 54 projects to mark. Technically this is sampling without replacement.

Answer 10

Ah okay, that makes sense.

Answer 11

For calculating (b) your proposal is correct.

Answer 12

Grr, that's tedious :P

For c). Would you multiply P(at least 10 in section 1) with (P at least 10 in section 2)?

Answer 13

Or is it add because it could be all in S1 OR all in S2

Answer 14

I used an online calculator for (b).
\[\large P(X \ge10)=0.517\]

Answer 15

Ah okay, that matched the answer they gave! :)

Answer 16

For (c), the event that at least 10 are from the first section and the event that at least 10 are from the second section are mutually exclusive. Therefore addition of the probabilities of these two events would seem to be the right way to go.

Answer 17

Using the online calculator again, gives 0.523 for the probability of (c).

Answer 18

That makes sense too! Thank you very much! :)

Answer 19

You're welcome :)