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Mathematics 16 Online
OpenStudy (anonymous):

A student faculty government committee of 4 people is to be formed from 20 student volunteers and 5 faculty volunteers. Find the probability that the committee will consist of the following, assuming the selection is made at random: a. two students and two faculty members and b. one faculty member and three students.

OpenStudy (anonymous):

a) (20*19*5*4)/(20*19*18*17+ 20*19*18*5 + 20*5*4*3 + 5*4*3*2) b) (5*20*19*18)/(20*19*5*4 + 20*19*18*17 + 5*4*3*2 + 5*4*3*20)

OpenStudy (anonymous):

Part a: So from a group of 5 faculty you will chose two members, and from a group of 20 students you will chose two members. The probability is the number of different ways you can choose 2 faculty and 2 students divided by the total number of ways you can pick 4 people from 25. \[\frac{{5 \choose 2}{20 \choose 2}}{25 \choose 4} = \frac{10\times 190}{12650} \approx 15\%\]

OpenStudy (anonymous):

Thank you!

OpenStudy (anonymous):

what about part b? Would I follow the same process as above? 5C1 x 20C3 / 25C4?

OpenStudy (anonymous):

Exactly.

OpenStudy (anonymous):

Do you get 0.090?

OpenStudy (anonymous):

Nope.

OpenStudy (anonymous):

What did I do wrong?

OpenStudy (anonymous):

I got it!

OpenStudy (anonymous):

Should be 45%.

OpenStudy (anonymous):

I had typed something into my calculator incorrectly but have the same answer. Many thanks for making it easier to understand.

OpenStudy (anonymous):

Happy to help! =)

OpenStudy (anonymous):

Can you help me with another similar problem?

OpenStudy (anonymous):

Here it is: Fred needs to print his term paper. He pulls down a box of 12 ink cartridges and recalls that 3 of them have no more ink. If he selects 4 cartridges from the box, find the probability that a) one cartridge has no ink and b) three cartridges have no ink. These are hard for me with all the different ways they can word them. Thanks :>)

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