Must be Bayes' theorem A professor gave me the following question a week earlier that he himself was doubtful about. I gave it quite a lot of thought but couldn't come up with any way to solve the question. Here it is:-
$Q.$A wallet containing $16$ currency notes - $4$ each of $Rs. 10$, $Rs. 20$, $Rs. 50$ and $Rs. 100$ - is kept on a table in a dark room. The maid of the house steals $2$ notes from the wallet which are apparently not visible to her (i.e. she doesn't know which notes she has taken) and leaves the room. After some time the owner of the wallet comes into the room and draws $2$ notes from the wallet. After coming out of the room, he finds that the amount in his hand is $Rs. 200$. So find the probability that the notes the maid has stolen comprises of two $Rs. 100$ notes.
During discussion of this question with my friends one of them said that since two $Rs. 100$ notes are surely left, the probability should be $$P=\frac {\binom 42}{\binom {14}2}=\frac 6{91}.$$ But the answer given is $\frac 1{91}$. I think that to solve this question one will have to use Bayes' theorem. But where and how should I use it?
 A: There is no need to apply Bayes Rule. The numerator of your expression is incorrect. If the owner has $2$ Rs $100$ notes, then there is only one way for the maid to take two $100$ Rs notes. So the numerator is not $\binom{4}{2}$ but $1$.
Extra: I'd like to point out that in this problem, the maid taking two notes and the owner taking two notes are reversible.
A: To put it another way: The owner took two $100$ bills.  That means there were $14$ other bills left for the maid to have taken previously, and thus $_{14}C_2 = 91$ different ways for her to have chosen bills from the wallet.  Only one of those ways also yields two $100$ bills, so the desired probability is $1/91$.
A: The approaches suggested earlier appear more efficient, but I think you are right in the sense that you can also solve it with the Bayes theorem, e.g. as suggested below:
(I try to follow the terminology used at this source:  http://www.randomservices.org/random/dist/Conditional.html)
Define the set of notes stolen by the maid as discrete random variable $A_i$ (all possible realizations indexed $1,.,i,..n$) and the set of notes taken by the owner as event (or information) $B_i$.
$A_1$ is the realization with the maid getting two 100 Rs bills.
$B_1$ is the information that the owner has two 100 Rs bills.
You are looking for the conditional (posterior) probability that the maid stole 200 Rs, given the information that the owner has 200 Rs, i.e. you are looking for:
$P(A_1|B_1)$
Write the Bayes theorem as follows:
$P(A_1|B_1) = \cfrac{P(B_1|A_1)P(A_1)}{P(B_1|A_1)+P(B_1|A_2)+...+P(B_1|A_n)}=\cfrac{P(B_1|A_1)P(A_1)}{P(B_1)}$ 
Note that here the unconditional (prior) probability that the maid takes 200 Rs $P(A)$ equals the unconditional probability that the owner has 200 Rs, i.e. $P(A_1)=P(B_1)$.
Hence you know:
$P(A_1|B_1) = P(B_1|A_1)$
The owner (like the maid) draws twice, one note each, after the maid has stolen two times 100 RS, the probability to get 100 with the first draw is 2/14 and with the second draw is 1/13, i.e.:
$P(B_1|A_1)=\cfrac{1}{7*13}=\cfrac{1}{91}=P(A_1|B_1)$
A: I like Brian’s answer, but if you’re uncomfortable with the idea of counting the number of ways the maid can do something in the past, here’s another way to look at it.
Call the bills the maid took $b_1$ and $b_2$, and the ones the owner took $b_3$ and $b_4$. The question is then this: Given an arrangement $b_1, b_2,\dots,b_{16}$ of the bills where $b_3=b_4=100$, what is the probability that $b_1=b_2=100$? 
Equivalently, of all possible arrangements of these 14 bills: four $10$s, four $20$s, four $50$s, and the two $100$s that are not in positions $3$ and $4$, what fraction have the two $100$s first?
There are $\dfrac{14!}{4!4!4!2!}$ arrangements of the bills, and the number of them with the two $100$s first is the number of arrangements of four $10$s, four $20$s, four $50$s, or $\dfrac{12!}{4!4!4!}$, so the answer is
$$\left.{\dfrac{12!}{4!4!4!}}\middle/{\dfrac{14!}{4!4!4!2!}}\right.=\frac{12!2!}{14!}=\frac{1}{91}.$$
