# Probability that all three will buy the same donut

I saw the question below on a Q/A site,

Donut Empire sells five different types of donuts. Amanda finds herself 3rd in line in the queue behind her friends Nicole and Brenda and realises that there are only three of each type left.

a) Given that Nicole and Brenda only buy one donut each, find the probability that all three of them will buy the same donut. (No: You may assume that Nicole, Breda and Amanda are equally likely to buy any of the donuts available.)

Now my answer came out to be 1/3, as probability= (choices to look for)/(total choices)= 5/15

But the OP stated that the answer was (1/91)^3 Can anyone explain why?

Thanks.

-
I don't understand why we should spend time thinking about your questions if you don't even spend the minimal time it would take to notice that there's a typo in your answer that renders it incomprehensible. –  joriki Oct 21 '11 at 11:11
@joriki: Sorry about that. I have fixed it. –  Fahad Uddin Oct 21 '11 at 11:19
Let's say Nicole is first in line. No matter what kind she buys, what is the probability that Brenda buys the same kind? And given that Brenda bought the same kind as Nicole, what is the probability Amanda gets the same kind? (By the way, the answer $(1/91)^3$ is wrong. For one thing, it is unreasonably small.) –  André Nicolas Oct 21 '11 at 11:38

The probability that all three of them will buy the same donut is zero. Donut shops usually respect property laws and don't sell things they've already sold to other customers.

Assuming that the question was intended to ask about the probability that all three of them will buy the same type of donut, the question arises whether the sentence "You may assume that Nicole, Breda and Amanda are equally likely to buy any of the donuts available." is intended as written, or is also intended to refer to types of donuts.

If it's intended to refer to types of donuts, the answer is $1$ in $25$: There are $5^3$ combinations for the five types of donuts, they're all equally likely, and in $5$ of them they all buy the same type of donut. Since we wouldn't need to know how many of each type there were left to answer this, this is presumably not how the question was intended.

So assume that the question was intended to ask about the probability that all three of them will buy the same type of donut, given that they are equally likely to buy any of the donuts available. In that case, there are $\binom{15}3$ combinations of choices, all of which are equally likely, and there are again only $5$ combinations in which they all buy the same type of donut, so the probability in this case would be

$$\frac5{\binom{15}3}=\frac5{455}=\frac1{91}\;.$$

-

Imagine that Nicole goes first, and gets a some kind of doughnut (say chocolate, but what she gets doesn't matter).

The probability that Brenda gets the same kind as Nicole is $\frac{2}{14}$, since exactly $2$ of the remaining $14$ doughnuts are of the right kind. And given that Brenda gets the same kind as Nicole, the probability that Amanda gets the last doughnut of this kind is $\frac{1}{13}$. Thus the probability all three customers get the same kind is $$\frac{2}{14}\cdot \frac{1}{13}.$$

Comment: One could also do a "counting" version of the idea. After Nicole has chosen, Brenda has $14$ choices, and for every choice Brenda makes, Amanda has $13$ choices, for a total of $14\cdot 13$. All these choices are equally likely. We now count the choices in which all three get the same kind. there are $2$ ways for Brenda to get the same kind as Nicole did, and once Brenda has chosen, there is only $1$ way that Amanda can follow suit, for a total of $2\cdot 1$. The probability is therefore $\frac{2\cdot 1}{14\cdot 13}$.

-