How to combine probablities in the following example? I made the following example to understand more about combining probabilities.
Let's say we have 3 coins hidden inside 11 boxes (3 boxes each have 1 coin and the other 8 have 0 coin). So the probability of finding a coin inside a box is 3/11.
Someone give us additional information which is that the right 10 boxes have 2 coins and left 2 boxes have 1 coin hidden in them as shown in the image.
My question is what is the probability of finding a coin in box number 2 (The intersecting box). Intuitively, one expects that the probability of finding a coin in the intersecting box would be more than 2/10 and less than 1/2, but I fail to estimate and formalize that.
Sorry if this is trivial but I couldn't figure a convincing answer by myself. Your help is deeply appreciated.
Edit: After thinking more about it. I realize that the intersecting box should have 0 coin because that is the only way to satisfy the additional given information. However, my question is still how to generalize this intuition into an equation?


If e.g. the sets $A,B,C$ form a partition of index set $[11]$ then you must be able to find expressions for probabilities like: $$P(X_A=a,X_B=b,X_C=c)$$ where $n:=a+b+c$ equals the total number of coins (in your case $n=3$) and where for every $S\in\{A,B,C\}$ random variable $X_S$ denotes the number of coins that are placed in boxes that have an index in $S$.
Applying multi hypergeometric distribution we find:$$P(X_A=a,X_B=b,X_C=c)=\frac{\binom{|A|}{a}\binom{|B|}{b}\binom{|C|}{c}}{\binom{11}{n}}$$
In that context you can also find conditional probabilities.

Edit:
As an example suppose that you want to find the probability that box 2 hides a coin, but this time under the additional info that (again) the right $10$ boxes have $2$ coins but now (slightly different then your own example) the left $3$ boxes have $2$ coins. This again with a total of $11$ coins.
Then we go for finding:
$$P\left(X_{\left\{ 2\right\} }=1,X_{\left\{ 1,2,3\right\} }=2,X_{\left\{ 2,\dots,11\right\} }=2\right)=$$$$P\left(X_{\left\{ 1\right\} }=1,X_{\left\{ 2\right\} }=1,X_{\left\{ 3\right\} }=0,X_{\left\{ 4,\dots,11\right\} }=1\right)=$$$$\frac{\binom{1}{1}\binom{1}{1}\binom{3}{0}\binom{8}{1}}{\binom{11}{3}}$$
and:
$$P\left(X_{\left\{ 1,2,3\right\} }=2,X_{\left\{ 2,\dots,11\right\} }=2\right)=$$$$P\left(X_{\left\{ 1\right\} }=1,X_{\left\{ 2,3\right\} }=1,X_{\left\{ 4,\dots,11\right\} }=1\right)=$$$$\frac{\binom{1}{1}\binom{2}{1}\binom{8}{1}}{\binom{11}{3}}$$
(Note that both the second expressions involve partitions)
telling us that:
$$P\left(X_{\left\{ 2\right\} }=1\mid X_{\left\{ 1,2,3\right\} }=2,X_{\left\{ 2,\dots,11\right\} }=2\mid\right)=\frac{8}{16}=\frac{1}{2}$$
I admit that this is cumbersome, but my answer was focused on a more general situation (you wanted to generalize your intuition into an equation). Actually the essence of it is the formation of suitable partitions.
A shortcut is concluding that under the conditions we must have $X_{\{2,3\}}=1$ and that the boxes 2 and 3 have equal chance to be the one that has the coin.
