Question

There are 9 Republicans and 5 Democrats on a committee. A sub committee of four is selected at random. What is the probability that all are Republicans, given that at least two are Republicans?

Answer 1

yeah,, but the teachers love to give out these kind of question..

Answer 2

Poorly worded question.

Answer 3

First find the probability that two are republicans.

Answer 4

Well we have 9 republicans, and there is a probability of there being four of them. All we have to do is put 4 on top of the total of republicans and Democratsand we have our solution:

9 + 5 = 14
4/14

Now we simplify:

4/14
2/7
There is a 2/7 chance of there being all republicans chosen.

Answer 5

Then find the probability that 4 or republicans.

Answer 6

Ah, I didn't see that last part.

Answer 7

Well we just take away two from the total:

4/12
2/6
1/3

There is a 1/3 chance of there being all republicans.

Answer 8

Okay the probability that at least 2 are republicans is a bit more complicated.

Answer 9

Its the probability that 2, 3, 4 republicans are in it.

Answer 10

So for exactly 2 it is:\[
\frac{9}{14}\times\frac{8}{13}\times\frac{5}{12}\times\frac{4}{11}=\frac{60}{1001}
\]For exactly 3 it is: \[
\frac{9}{14}\times\frac{8}{13}\times\frac{7}{12}\times\frac{5}{11}=\frac{5}{143}
\]For exactly 4 it is: \[
\frac{9}{14}\times\frac{8}{13}\times\frac{7}{12}\times\frac{6}{11}=\frac{18}{143}
\]
Add all of these up.\[
\frac{60}{1001}+\frac{5}{143}+\frac{18}{143}=\frac{17}{77}
\]
The conditional probability is just the probability of exactly 4 AND at least 2 divided by the probability of at least two.

At least 2 and exactly 4 is not different than exactly 4 because exactly 4 implies at least 2. So we have: \[
\frac{18}{143}\div\frac{17}{77}=\frac{126}{221}\approx 57\%
\]

Answer 11

A tedious question indeed.