Assume that there are 7 different issues of Time, 6 different issues of Sports Illustrated, and 4 different issues of Newsweek, including the December 1st issue, on the rack. You choose 4 of them at random. (1) What is the probability that you choose 2 issues of Time and 2 issues of Sports Illustrated? (2) What is the probability that you choose at least 3 of the Sports Illustrated magazines?

Question

Assume that there are 7 different issues of Time, 6 different issues of Sports Illustrated, and 4 different issues of Newsweek, including the December 1st issue, on the rack. You choose 4 of them at random. (1) What is the probability that you choose 2 issues of Time and 2 issues of Sports Illustrated?    (2) What is the probability that you choose at least 3 of the Sports Illustrated magazines?

directrix · Answer

@glacier. I think this is the same problem worked on Open Study previously. Check and see. http://openstudy.com/updates/4f2b526fe4b039c5a5c87c99

anonymous · Answer

it is similar but the numbers are different i could figure out 2 on my own but I am still unsure about 1

directrix · Answer

1) You want 2 of the 7 Time magazines --> C(7,2)  Read as: Seven, choose 2

2) You want 2 of the 6 SI --> C(6, 2)

3) You want 0 of the 4 N --> C(4,0)

directrix · Answer

1) What is the probability that you choose 2 issues of Time and 2 issues of Sports Illustrated?

[C(7,2) * C(6, 2) * C(4,0) ] / C(17,4)  = 9/68 = .132 approx
       
Note: For the denomininator, there are 17 mags and you want to know how many ways any 4 can be chosen.

anonymous · Answer

thank you it makes sense but 2 still confuses me

directrix · Answer

(2) What is the probability that you choose at least 3 of the Sports Illustrated magazines? 

On this you will have four cases: 3 SI, 4 SI, 5 SI, and 6 SI.

See if you can figure out this one.

anonymous · Answer

[C(6,3) C(11,1)] / C(17,4)?

directrix · Answer

Yes for Case One.

directrix · Answer

Case Two - 4 SI

[C(6,4) * C(11,0)] / C(17,4)

Is this correct?

anonymous · Answer

I don't think you can choose 0 of 11 or am i wrong because what would you do in case 3 where you would need [C(6,5) * C(11,?)] / C(17,4)

directrix · Answer

C(11,0) = 11! / ( (11-0)! * 0!) = 11! / 11! 1!  = 1

0! by definition = 1.  The C(11,0) is really just a place holder of sorts to remind the solver that none of other other 11 magazines are to be chosen.

directrix · Answer

There is no case 3 and no case 4, I just realized. We are only choosing 4 magazines. So, part 2 is this: [C(6,3) C(11,1)] / C(17,4) + [C(6,4) * C(11,0)] / C(17,4) = 47/476 = .0987 approx Check out the calculation: http://tinyurl.com/mb2htrc

anonymous · Answer

okay :) I get it now thank you so much for helping me especially because it is 4am lol have a great rest of the night or day!