Question

In a cookie jar, there are 9 chocolate chip cookies and 13 sugar cookies. A hungry boy takes out the cookies one at a time and eat them. What is the probability that the 10th cookie that a boy eats is a chocolate chip cookie?

Answer 1

lol how did you even get that?

Answer 2

I can try to help, I'm thinking up a model right now.

Answer 3

How about we lock the chocolate chip cookie in the 10 position and figure out how many ways to assign the cookies in the remaining spots? Does that seem sound?

Answer 4

So we'd have 8 chocolate cookies and 13 sugar cookies, and we're assigning them among the 21 remaining spots.

Answer 5

If I recall, you do the progrssive probablities of each event, then multiply them. But stats was a while back. I am sure I could look it up.

Answer 6

Progressive probability would take me forever to find the answer.

Answer 7

Okay so here is my model @geerky42

We start out with 9c and 13s.
We give each cookie a number from 1 to 22.
He eats the cookies in the order of the number the get.

1) Assign a chocolate cookie the number 10.
\[
1 \text{ way to do this}
\]
Now we have 8c and 13s and 21 numbers.
Order no longer matters though.

2) Assign the remaining chocolate cookies to their numbers
\[
\binom{21}{8} \text{ ways to do this}
\]

3) The put sugar cookies in the remaining spots.\[
1 \text{ way to do this}
\]

So we multiply together to get: \[
1\times \binom{21}{8} \times 1 = \binom{21}{8} =\frac{21!}{8!13!}
\]

Answer 8

See any flaw in this model? It'd be easier for you to spot the flaw than for me to find it.

Answer 9

Ah, yes. Factorials. That should get the same results. Like I said, been a while. That is, if I recall, what the progressive examples lead to.

Answer 10

I don't see any flaw. I don't know.

Answer 11

Do you understand it at least?

Answer 12

Yes.

Answer 13

I think.

Answer 14

I used combinations to give the chocolate cookies numbers because chocolate cookies are indistinguishable among each other. Since sugar cookies are indistinguishable among each other, there was only one way to assign them the remaining numbers.

Answer 15

Yeah, I understand, but how can this help me with my current problem? I don't get your point.

Answer 16

And as for the assignment of 10 to a chocolate cookie, since the chocolate cookies are indistinguishable there is only one way to do it.

Answer 17

Well it is the total number of 'target' outcomes.

Answer 18

For the number of all outcomes... we modify it slightly.
1) Assign the remaining chocolate cookies to their numbers
\( \Large \binom{22}{9} \) ways to do this.

2) The put sugar cookies in the remaining spots.
1 way to do this.

So the total ways is: \[
\binom{22}{9}\times 1 = \binom{22}{9} =\frac{22!}{9!13!}
\]

Answer 19

The probability of the target event happening is: \[\Large
\Pr(A) =\frac{|A|}{|\Omega|} = \frac{\frac{21!}{8!13!}}{\frac{22!}{9!13!}}
\]

Answer 20

Hmmmmm\[
\Large
\Pr(A) =\frac{|A|}{|\Omega|} = \frac{\frac{21!}{8!13!}}{\frac{22!}{9!13!}} = \frac{\frac{21!}{8!13!}}{\frac{22(21!)}{9(8!13!)}} = \frac{1}{\frac{22}{9}}=\frac{9}{22}
\]

Answer 21

Ohh. I get it. Thank you so much