The number of colds person gets in a year has ${\rm Poisson}(3)$ distribution. A new drug lower it to ${\rm Poisson}(.75)$ and is effective for $8$ out of $10$ people. The entire population was given the drug. One person was selected at random from the population and was found to have $1$ cold during the year. What is the probability that the drug was effective for this person?
Assuming $A$ is having $1$ cold per year and $B$ the drug is effective.
By knowing $P(B) = 8/10$ and finding the probability of having 1 cold/year before and after the drug. $P(A|B) = \exp(-.75)(.75)=.354275$ and $P(A|B^{c}) = exp(-3)(3)= .14936$, I can almost use the Bayes' theorem to find the $P(B|A)$ which is the probability the drug being effective knowing the person have been chosen had only 1 cold.
$P(B|A) = \frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|B^{c})P(B^{c})}$.
I still miss a part of the equation $P(B^{c})$ if my assumption was correct.
The mistake is in $$P(A\mid B^c)=\exp(-3)3=0.149\neq 0.22245$$ as you have it. So this is only a calculation mistake, rest including $P(B^c)=0.2$, is correct. To see this note that there are only two outcomes (as defined):
These are by their definition complementary. $B$ occurs or $C$ occurs. So, it must be that $P(B)=1-P(C)$ (actually $C$ stands for $B^c$ here).