cant figure this out at all.

basically there are 5 cds that come in a cereal box randomly. what are the chances of getting all 5 cds if you buy 12 boxes?

can someone point me in the right direction?

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- April 21st 2011, 05:53 AMBaffzword problem
cant figure this out at all.

basically there are 5 cds that come in a cereal box randomly. what are the chances of getting all 5 cds if you buy 12 boxes?

can someone point me in the right direction? - April 21st 2011, 12:11 PMpoirot
You need more information, i.e. the probability of getting a cd

- April 21st 2011, 02:27 PMPlato
This is an

*inclusion/exclusion*question.

I deleted one reply because the notation was too complicated.

Lets say the CD's are .

Letbe the event that at least one disk*X*is present in a sample of 12.**x**

Now we want to show http://quicklatex.com/cache3/ql_96c0...d8f7d1f_l3.png

That is http://quicklatex.com/cache3/ql_9162...289b3a6_l3.png

Note that http://quicklatex.com/cache3/ql_807c...5c6672e_l3.png - April 22nd 2011, 11:49 AMArchie Meade
If the cereal box is guaranteed to have a CD in it

and each of the CDs have equal probability of being placed in a cereal box,

then you can find the probability of NOT finding all five CDS in the 12 boxes

by calculating the probabilities that

(1) one of the CDs is not in the 12 packets

(2) two of the CDs are not in the 12 packets

(3) three of the CDs are not in the 12 packets

(4) four of the CDs are not in the 12 packets

All 5 CDs cannot be missing of course.

The probability that one CD is missing is 5C1(4/5)^(12)

because one of the other 4 CDs are in each box for each missing one.

The probability that two CDs are missing is 5C2(3/5)^(12)

because one of the other 3 CDs is in each box for every missing pair.

The probability that three CDs are missing is 5C3(2/5)^(12)

Also, include four CDs missing.

Now the question is...

Does the probability of one being missing include the probability of 2, 3 or 4 also missing ?

Subtract the probability of at least one CD missing from 1.