Question help!!!

VijaKhara@gmail.com · science forum beginner Joined: 30 Sep 2005 Posts: 26

My solution is wrong since it is different from book's answer. But I
cannot find out what point I missed.

Given 12 chocolate cookies there are 7 chocolate chips that are
randomly placed into the cookies. What is the probability that at least
one cookie has at least two chips?

My solution:

P("at least one cookie has at least two chips")=1- P(there is no cookie
which has more than 1 chip")=1-P(all cookies have either 1 chip or zero
chip).

I can think of 12 cookies as 12 urns:

| | | | | | | | | | | | |

and try to arrange 7 chocolate chips into these urns such that each
chip will be put into a different urn:

The number arrangements will be 12C7
The total possible arrangements is 12^7

Such P=1-12C7/12^7.

Thanks

Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

VijaKhara wrote;
"Given 12 chocolate cookies there are 7 chocolate chips that are
randomly placed into the cookies. What is the probability that at least
one cookie has at least two chips?
My solution:
P("at least one cookie has at least two chips")=1- P(there is no cookie
which has more than 1 chip")=1-P(all cookies have either 1 chip or zero
chip).
I can think of 12 cookies as 12 urns:
| | | | | | | | | | | | |
and try to arrange 7 chocolate chips into these urns such that each chip
will be put into a different urn:
The number arrangements will be 12C7
The total possible arrangements is 12^7
Such P=1-12C7/12^7."
______________________________________
Re;
It looks like you are attacking the problem in right way by first
determining the probability that all chips fall on virgin cookies then
subtracting this from one. I would attack it like this: The
probability that the first chip falls on a virgin cookie is 12/12 or 1;
for the second chip it is 11/12; for the third, 10/12 and so on. The
probability then that all 7 chips fall on virgin cookies is then
12!/(5!)(12^7)=0.111 or 11.1%. Thus, the probability of anything other
than this (which seems to satisfy your query) is simply 1-0.111=0.889 or
88.9%. Does that agree with your book?

-Dan Akers

VijaKhara@gmail.com · science forum beginner Joined: 30 Sep 2005 Posts: 26

Hi Akers, your result is the same as the book but why my result is
wrong.
Is the total of arrangment such that no cookie has more than two chips
is 12C7?? thanks

Dan Akers wrote:

Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

Vija Kara wrote;
"Your result is the same as the book but why my result is wrong.
Is the total of arrangement such that no cookie has more than two chips
is 12C7?"
____________________________________
Re:
You are making a very common mistake. When the set of 7 chips are
distributed on the set of 12 cookies, with one chip/cookie, you are not
dealing with combinations, but rather permutations. Thus, the
probability of one chip per cookie in this problem is 12P7/12^7. And
the answer to the problem is then the probability of anything else that
could occur with the chips on cookies, namely: 1-12P7/12^7

-Dan Akers

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