probability question...first correct answer gets 5 pts

[pico]

there is a game that i played last night. there are 9 identical cards. 4 of them has a heinekin picture in the back, the remaining 5 cards has nothing. you get to pick 4 cards at random. and the prizes are:

no beer card = one heinekin coaster
1 beer card = one pint of beer plus a pack of coasters
2 beer cards = 3 pints of beer
3 beer cards = 5 pints of beer
4 beer cards = 5 pints of beer plus a mini heinekin refrigerator

question is, what are the probability of getting no beer card, 1 beer card, 2 beer cards, 3 beer cards, and 4 beer cards.

the odds for 4 beer cards is 1/126, which is about 0.8%

[Shortstop]

Picochalk

[RonPaul2008]

LOL somebody wants their homework done for them

[Lazyboy311]

I'll take it:

All conditional probability, individual chance % listed in yes/yes/...no format. Multiplied by number of possible sequences within each outcome. All numbers rounded and results in a .2% rounding error.

0 beer card: .555 x .5 x .428 x .33 = .0395 x 1 = .040 = 4%
1 beer card: .444 x .625 x .571 x .5 = .0792 x 4 = .32 = 32%
2 beer cards: .444 x .375 x .714 x .666 = .0784 x 6 = .47 = 47%
3 beer cards: .444 x .375 x .286 x .833 = .039 x 4 = .16 = 16%
4 beer cards: .444 x .375 x .285 x .166 = .0079 x 1 = .008 = 0.8 %

[THEGREAT30]

5 points huh

[Ganchrow]

I'll take it:

All conditional probability, individual chance % listed in yes/yes/...no format. Multiplied by number of possible sequences within each outcome. All numbers rounded and results in a .2% rounding error.

0 beer card: .555 x .5 x .428 x .33 = .0395 x 1 = .040 = 4%
1 beer card: .444 x .625 x .571 x .5 = .0792 x 4 = .32 = 32%
2 beer cards: .444 x .375 x .714 x .666 = .0784 x 6 = .47 = 47%
3 beer cards: .444 x .375 x .286 x .833 = .039 x 4 = .16 = 16%
4 beer cards: .444 x .375 x .285 x .166 = .0079 x 1 = .008 = 0.8 %

Yep. Well done.

For those still a bit quizzical as to how such an answer might be quickly obtained in general:

Underlying this problem is the Hypergeometric distribution. In Excel this is represented by the =HYPGEOMDIST() function.

0 cards => =HYPGEOMDIST(0,4,4,9) = 5 126 ≈ 3.97%
1 cards => =HYPGEOMDIST(1,4,4,9) = 40 126 ≈ 31.75%
2 cards => =HYPGEOMDIST(2,4,4,9) = 60 126 ≈ 47.62%
3 cards => =HYPGEOMDIST(3,4,4,9) = 20 126 ≈ 15.87%
4 cards => =HYPGEOMDIST(4,4,4,9) = 1 126 ≈ 0.79%

In general, the probability of selecting exactly w winners out of n draws without replacement given an N sized population consisting of W winning draws (where 0≤w≤W≤N and 0≤w≤n≤N) is Excel given in Excel by:

= HYPGEOMDIST(w,n,W,N)

Using combinatorials this may be equivalently expressed as:

= combin(W, w) * combin(N-W, n-w) / combin(N,n)