"Albert owns 5/9ths of the stock in the North West Chocolate Company. His sister, Rena, owns half as much stock as Albert. What part of stock is owned by NEITHER Albert nor Rena?"
The answer is obviously 1/6 (or 3/18). However, a friend reasoned: Albert doesn't own 4/9ths and Rena doesn't own 13/18ths, so together they do not own 21/18ths or 1 1/6th. Obviously, that "one" does not belong - Where did that extra 1 come in and how can I, in a mathematically sound way, get rid of it?
Thanks!
Think about it like this, say person a owns $x$ and person b owns $y$ and say that there is $z$ not owned by either of them. Then $x+y+z=1$ and you can solve for $z$ as you did on your first attempt.
For your second attempt, the total that person a doesn't own is actually $1-x$ or $y+z$ and the total that person b doesn't own is $1-y$ or $x+z$ adding these two together gives us $x+y+2z=(x+y+z)+z=1+z$ since as mentioned before, $x+y+z=1$. Thus the extra 1 comes from the fact that you are counting the amount that person a owns, the amount that person b owns and then double-counting the amount neither of them own.
If you want to read more on these types of situations, you may wish to look up the "Principle of Inclusion-Exclusion" which gives a way to calculate these types of things in even more general settings.
Interesting question (+1) A rather non arithmetic approach: Shares not owned by Albert could be owned by his sister, and shares not owned by his sister could be owned by Albert. In other words, in your second approach, there are "not owned" shared counted twice. That's why the probability "exceeds" 1
I think it's a bit tricky to articulate where the extra $1$ came from and how to get rid of it since it originated from a combination of two separate errors.
Let $A$ be the set of shares Albert owns, let $R$ be the set of shares Rena owns, and let $U$ be the set of all shares. The set of shares Albert doesn't own is $A'=U\setminus A$; the set of shares Rena doesn't own is $R'=U\setminus R$.
The straightforward computation that you give first, stated in this language, is $$ \lvert U\setminus(A\cup R)\rvert=\lvert U\rvert-\lvert A\rvert-\lvert R\rvert=1-\frac{5}{9}-\frac{1}{2}\cdot\frac{5}{9}=\frac{1}{6}, $$ where the right side of this equation is justified by the properties that $A$ and $R$ are disjoint subsets of $U$.
The correct way to do the computation using the complements $A'$ and $R'$ is to compute $\lvert A'\cap R'\rvert$. I think your friend made two errors,
computing $\lvert A'\cup R'\rvert$ rather than $\rvert A'\cap R'\rvert$, and
computing $\lvert A'\cup R'\rvert$ using $\rvert A'\cup R'\rvert=\lvert A'\rvert+\lvert R'\rvert$, which ignores the overlap of $A'$ and $R'$.
Let's fix (2) first by using the principle of inclusion-exclusion, which states that $$ \lvert X\cup Y\rvert=\lvert X\rvert+\lvert Y\rvert-\lvert X\cap Y\rvert. $$ Applying this to the union of $A'$ and $R'$ gives $$ \lvert A'\cup R'\rvert=\lvert A'\rvert+\lvert R'\rvert-\lvert A'\cap R'\rvert=\frac{4}{9}+\frac{13}{18}-\lvert A'\cap R'\rvert=\frac{7}{6}-\lvert A'\cap R'\rvert. $$ How do we fix (1)? In other words, how do we compute $\lvert A'\cap R'\rvert$? Well, we can use the equation above! A bit of algebra gives $$ \lvert A'\cap R'\rvert=\frac{7}{6}-\lvert A'\cup R'\rvert. $$ This might not seem to be progress because it expresses the unknown intersection in terms of the unknown union. But actually the union isn't unknown: every share is unowned by one of Albert and Rena. So $A'\cup R'$ equals $U$, the set of all shares. Hence $$ \lvert A'\cap R'\rvert=\frac{7}{6}-\lvert A'\cup R'\rvert=\frac{7}{6}-\lvert U\rvert=\frac{7}{6}-1=\frac{1}{6}. $$
If I had to summarize, I'd say the extra $1$ comes from the union of the set of shares unowned by Albert and the set of shares unowned by Rena, which is, in fact, the set of all shares. The reason for subtracting this $1$ is to correct the two errors mentioned above.
