Note: I constructed this problem for learning purpose. I am not a student and this is no homework.
The Problem: Two persons 1, 2 have some valuable assets. They have access to 2 storage locations, location 1 and location 2. Person B is worried about possible robbery of his valuables. Person A is not concerned. Valuables of person B amounts to $x.
Person B wants to allocated his valuables in such a way to minimize total lost value for both persons in case of a robbery takes place.
What amount should person B use in each location so that the total loss of the 2 persons is minimized in case of robbery?
My Attempt:
$P1 = 0.30$ for Location $1$ is robbed
$P2 = 0.20$ for Location $2$ is robbed
$P3 = 0.10$ for Locations $1$ AND $2$ are robbed
$P4 = 0.40$ for No robbery will occur
Person A uses Location 2 Always.
Person B has 3 main options:
1-Use Location $1$ to Store Value $X$, Expected Loss:
$(0.3+0.1=0.4) X$
2-Use Location $2$ to Store Value $X$, Expected Loss:
$(0.2) Y$
3-Divide valuables between Locations $1$ and $2$ (Place $X-Y$ in one location and $Y$ in another location)
3.A: Place $X-Y$ in Location $1$ AND $Y$ in Location $2$, Expected Loss:
$(X-Y) * 0.3 + Y*(0.2)$ or Should this be $X*(P3=0.1)$?
3.B: Place $X-Y$ in Location $2$ AND $Y$ in Location $1$, Expected Loss:
$(X-Y)*0.2+(X-Y)*0.3$ or Should this be $X*(P3=0.1)$?
For options $3.A$ and $3.B$, $X$ has to be greater than or equal to $Y$ of course.
EDIT User YJT provided a good point which made me re-phrase the objective.
You can significantly simplify your calculations by using the linearity of the expectation. Let $X_i=1$ if location $i$ is robbed ($0$ otherwise). Suppose he puts $X$ in $1$ and $x-X$ in $2$. Then his expected loss is $XE(X_1)+(x-X)E(X_2)=X(E(X_1)-E(X_2))+const$.
Since $E(X_1)=0.3+0.1$ and $E(X_2)=0.2+0.1$, the loss is $0.1 X+const$. To minimize the loss, he should put $X=0$ in location 1.
This is very intuitive - to minimize the loss, put everything in the safer place. The probability that both are robbed is not important as it loses everything regardless of the split.