Finding probability distribution for a discrete random variable When the health department tested private wells in a county for two impurities commonly
found in drinking water, it found that 20% of the wells had neither impurity, 40% had
impurity A, and 50% had impurity B. (Obviously, some had both impurities). If a well is
randomly chosen from those in the county, find the probability distribution for the
discrete random variable Y = the number of impurities found in the well.
I already have the answer, and this isn't homework. Just trying to find out how to solve the problem.
 A: First, it is obvious that (by the inclusion-exclusion principle) there are $10\%$ of wells which have two impurities, which implies that there are $80\%-10\%=70\%$ of wells that have only one impurity. For $Y$ there are only three possible values: $0$, $1$, and $2$. Then the distribution of $Y$ is clearly
$$\begin{matrix}Y&|&0&1&2\\\mathbf P&|&0.2&0.7&0.1\end{matrix}.$$ 
A: Let $p_A$, $p_B$ and $p_C$ be the respective probabilities that a well contains impurity A only, impurity B only and both impurities. We know that
$$p_A+p_B+p_C+0.2=1.$$
Notice that the probability that a well contains impurity A is given by $p_A+p_C$ (and similarly for impurity B). Thus
$$p_A+p_C=0.4,\qquad p_B+p_C=0.5.$$
Now you just have a system of $3$ linear equations in $3$ unknowns. Is this something you can solve?
A: Given: $P(A)=0.4,P(B)=0.5,1-P(A\cup B)=0.2.$
Look at the Venn diagram:

$$\begin{cases}
P(A)=P(A\setminus B)+P(A\cap B)=0.4 \\ 
P(B)=P(B\setminus A)+P(A\cap B)=0.5 \\ 
P(A\cup B)=1-P(A\cup B)=1-P(A\setminus B)-P(B\setminus A)-P(A\cap B)=0.2
\end{cases} \stackrel{+}\Rightarrow $$
$$P(A\cap B)=0.1 \Rightarrow P(A\setminus B)=0.3, P(B\setminus A)=0.4.$$
Hence: 
$$\begin{align}
&P(Y=0)=1-P(A\cup B)=0.2 \\ &P(Y=1)=P(A\setminus B)+P(B\setminus A)=0.7 \\ &P(Y=2)=P(A\cap B)=0.1.
\end{align}$$
