Paddy Wants a Chocolate Chip Cookie

Let $n(A)$ be the number of cases where 7th cookie is a chocolate chip cookie. So we have to set 5 chocolate and 8 oatmeal, so $n(A)$ is $\binom{13}{5}$ . Let $n(S)$ be the total number of cases. We have to set 6 chocolate and 8 oatmeal, so $n(S) =\binom{14}{6}$ and the probability is $n(A)/n(S)=6/14=3/7$ so $a=3,b=7$ and $a+b=10$ .

The probability that of getting a chocolate chip cookie or an oatmeal cookie is evenly distributed among all n tries.

Then, the probability that Paddy picks a chocolate chip cookie on the 7th try is the same probability that Paddy picks a chocolate chip cookie on the 1st try.

The '1st try' has a probability of (no. of chocolate chip cookies)/(total no. of cookies.) = 6/(6+8) = 3/7

(a,b) = (3,7), therefore, a+b=10.

Notice the important observation that the chance of eating a chocolate chip cookie at every stage is equal, if we do not know what the type of cookies eaten before this stage.

Hence the probability for the 7th cookie is the same as that of the 1st cookie, hence the answer is $\frac6{14}=\frac37$ . Hence $3+7=10$ .

There are a total of 6+8= 14 cookies and 6 chocolate cookies. The probability of choosing a chocolate cookie is 6/14= 3/7.

The probability is the same for choosing a chocolate cookie in the 7th turn, in fact even in the last turn since we are not given any information about the previous selections.

Thus, the probability of choosing a chocolate cookie in the 7th turn=- 3/7.

Note that g.c.d(3, 7)= 1 so comparing we get a= 3 and b= 7 and a+b= 10.

It doesn't matter which cookie (i.e. 1st, 2nd, 3rd) it is, the probability that cookie is a chocolate chip, is still $\frac {6}{14} = \frac{3}{7}$ . Thus, the answer is $3+7 =10$ .

There are total number of 14 cookies. So the probability of the chocolate chip cookies taken is 6/14 . Simplify the fraction and we will get 3/7 . Now this fraction is commonly divisible only by 1, that means 3 and 7 are positive coprime integers. Therefore the value of a+b which is 3+7=10.