I throw a fair dice once. If it shows an odd number, I throw a fair coin that number of times. If dice shows an even number, I don’t throw the coin at all. What’s the probability that I get at least 2 heads?
I use "complementary events" approach. Let $A$ be the event that I get at least 2 heads. Then $A^{c}$ is the event I get less than 2 heads. Now $P(A) = 1 - P(A^{c})$. Let's choose partition of sample space $\{B_{k}\}$, where $k$ is the number shown on the dice when thrown, $k = 1,...,6$.
Then $P(A^{c}) = \sum_{k} P(A^{c}|B_{k})P(B_{k})$. For $k=2,4,6$, we have $P(A^{c}|B_{k})= 0$.
We find $P(A^{c}) = 1 \times \frac{1}{6} + \frac{4}{8} \times \frac{1}{6} + \frac{6}{32} \times \frac{1}{6} = \frac{9}{32}$,
giving $P(A) = \frac{23}{32}$. However direct approach yields $P(A) = \frac{7}{32}$.
Why does the "complementary events" method fail here?
For $k=2,4,6$, you have $P(A^c|B_k) = 0 ~$, which is incorrect. In fact for $k=2,4,6$, we have $P(A^c|B_k) = 1 ~$, as we do not toss the coin and have zero heads which is indeed less than two.
So, $ \displaystyle P(A^{c}) = 3 \times \frac16 + 1 \times \frac{1}{6} + \frac{4}{8} \times \frac{1}{6} + \frac{6}{32} \times \frac{1}{6} = \frac{25}{32}$
And that leads to $ \displaystyle P(A) = \frac{7}{32}$