Veany Eat chocolate chips In a cookie jar, there are 6 chocolate chip cookies and 8 oatmeal cookies. Veany takes out the cookies one at a time and proceeds to eat them. The probability that the 7th cookie that Veany eats is a chocolate chip cookie is $\frac{a}{b}$, where $a$ and $b$ are positive coprime integers. What is the value of $a+b$?
 A: The answer for $\frac{a}{b}$ is $\frac{3}{7}$, so $a + b$ is 10.
The first thing to do is denote Cs as chocolate chip cookies, and Os as oatmeal cookies. So we have $$CCCCCCOOOOOOOO$$ as our cookie jar. The problem asks for the probability that the 7th cookie is chocolate ship, and there's a variety of ways that could happen. The way to approach this problem is to come up with cases for the first 6 cookies, draw a slash denoting our 7 cookie as chocolate chip and then figure out the cases for the other 7 cookies.


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*For our first case we'll use all 5 chocolate chip cookies to the left of our "divider" (our 7th cookie). So it would look something like this $$OCCCCC|OOOOOOO$$ Now the ways we can arrange the cookies to the left of the divider is $6 \choose 1$, and the ways we can arrange the cookies to the right of the divider is $7 \choose 0$, so the total number of ways is ${6 \choose 1} \times  {7 \choose 0} = 6$

*Similarly for our second case we are going to have 2 oatmeal cookies to the left of the divider instead of 1 oatmeal cookie as in our previous case. So it would look like this $$OOCCCC|COOOOOO$$The ways we can arrange the cookies to the left of the divider is $6 \choose 2$, and the ways we can arrange the cookies to the right of the divider is $7 \choose 1$, so the total number of ways is ${6 \choose 2} \times  {7 \choose 1} = 105$

*So now for our third case we are going to have 3 oatmeal cookies to the left of our divider! That would look like this $$OOOCCC|CCOOOOO$$ The ways we can arrange the cookies to the left of the divider is $6 \choose 3$, and the ways we can arrange the cookies to the right of the divider is $7 \choose 2$, so the total number of ways is ${6 \choose 3} \times  {7 \choose 2} = 420$

*Are you noticing a pattern here? For every case, we have $n$ oatmeal cookies to the left of our divider, and $8-n$ cookies to the right of our divider. The ways we can arrange the cookies on the left side would be $6 \choose n$, and on the right side would be $7 \choose {n-1}$ where $n$ is a positive integer from 1 to 6. So our desired probability is just $${14 \choose 8} \times \displaystyle \sum_{i=1}^6 \bigg({6 \choose n} \times {7 \choose {n-1}}\bigg) = \frac{3}{7}$$

A: The simple approach is to realize the chance the 7th one is chocolate chip is the same as the first one, which is $\frac 37$.think of putting the cookies in order, then swapping the 1st and 7th.
A: All the permutations of the cookies are equally likely. So the probability is $\frac{6}{14}$, the same as the probability the first cookie is chocolate chip.
