[Combinatorics]
Problem: Oi-Lam tosses three fair coins and removes all of the coins that come up heads. George then tosses the coins that remain, if any. Determine the probability that George tosses exactly one head.

Solution: If Oi-Lam tosses 3 heads, then George has no coins to toss, so cannot toss exactly 1 head. If Oi-Lam tosses 2, 1 or 0 heads, then George has at least one coin to toss, so can toss exactly 1 head.

Therefore, the following possibilities exist:

* Oi-Lam tosses 2 heads out of 3 coins and George tosses 1 head out of 1 coin
* Oi-Lam tosses 1 head out of 3 coins and George tosses 1 head out of 2 coins
* Oi-Lam tosses 0 heads out of 3 coins and George tosses 1 head out of 3 coins

We calculate the various probabilities.

If 3 coins are tossed, there are 8 equally likely possibilities: $\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{TTH}$, THT, HTT, TTT. Each of these possibilities has probability $\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$. Therefore,



* the probability of tossing 0 heads out of 3 coins is $\frac{1}{8}$
* the probability of tossing 1 head out of 3 coins is $\frac{3}{8}$
* the probability of tossing 2 heads out of 3 coins is $\frac{3}{8}$
* the probability of tossing 3 heads out of 3 coins is $\frac{1}{8}$

If 2 coins are tossed, there are 4 equally likely possibilities: HH, HT, TH, TT. Each of these possibilities has probability $\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$. Therefore, the probability of tossing 1 head out of 2 coins is $\frac{2}{4}=\frac{1}{2}$.

If 1 coin is tossed, the probability of tossing 1 head is $\frac{1}{2}$.

To summarize, the possibilities are

* Oi-Lam tosses 2 heads out of 3 coins (with probability $\frac{3}{8}$ ) and George tosses 1 head out of 1 coin (with probability $\frac{1}{2}$ )
* Oi-Lam tosses 1 head out of 3 coins (with probability $\frac{3}{8}$ ) and George tosses 1 head out of 2 coins (with probability $\frac{1}{2}$ )
* Oi-Lam tosses 0 heads out of 3 coins (with probability $\frac{1}{8}$ ) and George tosses 1 head out of 3 coins (with probability $\frac{3}{8}$ )

Therefore, the overall probability is $\frac{3}{8} \cdot \frac{1}{2}+\frac{3}{8} \cdot \frac{1}{2}+\frac{1}{8} \cdot \frac{3}{8}=\frac{27}{64}$.

Final answer: $\frac{27}{64}$