Easy statistics problem question!

[GoldenBear]

Okay a guy has 6 rock and 5 rap CD's, what is the probability that he'll draw at LEAST 2 rap?

What I have so far is the prob of picking all rap plus the rap and 1 rock..

(5/11)(4/10)(3/9) + (5/11)(4/10)(6/9)

What else do I need?

EDIT: Forgot to mention he draws THREE CD's

 

[dopcombo]

ahh...

does he draw from the piles separately?


*edited for goldenbear's edit*

 

[Noriaki]

How many CDs is he drawing?

 

[GoldenBear]

Whoop, forgot to mention he only picks THREE CD's..

 

[Noriaki]

well if he's taking 3 CDs and 2 have to be rap, you just need the 2 cases where all 3 are rap or 2 rap + 1 rock.

5C3/11C3 + 5C2*6C1/11C3 I think.

5C3 is the number of ways to pick 3 rap CDs, 11C3 is the total number of way to pick 3 CDs.

5C2*6C1 is 2 Rap from 5 and 1 Rock from 6. Again 11C3 is the total.

(If you want to be really picky the first one should be 5C3*6C0/11C3, but xC0 = 1 for all x).

 

[HomeBrewerDude]

You didn't indicate how many CD's will be picked. If this isn't specified, you have to compute the probability distribution for the total number of draws AS the number of draws varies (or at least the meaningful part of the distribution).

To draw at least two Rap, you must select at least 2 of the 11 CDs and you cannot not draw anymore than 11 b/c you only have 11 CDs.

When n=2
RAP RAP
p(@least2RAP)=5/11*4/10

When n=3
RAP RAP RAP
RAP RAP ROCK
RAP ROCK RAP
ROCK RAP RAP
p(@least2RAP)=(5/11*4/10*3/9)+3(5/11*4/10+6/9)

so forth until n=7 at which point the P(@least2Rap)=1.0
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The way you are trying to solve the problem suggests that the guy is only going to take 3 CD's. IF this is true then you are on the right track. However there is more than one combination for picking 2 rap and 1 rock.

 

[GoldenBear]

Wow you're a GEENUS.

And thanks to dop too for working through my mistake =)

 

[HomeBrewerDude]

Noriaki and me should get the same answer, though through different arithmetic.

 

[Noriaki]

Disclaimer: It's been nearly 2 years since I took my probability course (which is what this is by the way, it's not Stats. Stats&Prob is usually in one course at least for the intro and they are related, but they are very different).

yamahaXS: Of course, it works out to the same thing if you simplify all the factorials in those Chose functions I used, I'm just lazier than you and since most caluclators have the Choose function it's pretty simple