Insects of a certain type lay eggs on leaves such that the number of
eggs on a given leaf has a Poisson distribution with mean $1$. For any
given leaf, the probability is $0.1$ that it will be visited by such
an insect, and leaves are visited independently of each other.(a) What is the probability that a given leaf has no eggs?
(b) If a leaf is inspected and has no eggs, what is the probability that it has been visited by an insect?
(c) If 10 leaves are inspected and none have any eggs, what is the probability that at least one leaf has been visited by an insect?
Let $X=\{\text{Number of eggs on a given leaf}\}$ and we have that $X\sim \text{Poi}(\lambda), \ \lambda=1.$ We also know that $P(\text{Visit}) = 0.1.$ We are looking for $P(\text{No eggs}).$
So, according to Baye's theorem: $$P(\text{No eggs})=1-P(\text{Eggs})=1-P(\text{Eggs|Visit})\cdot P(\text{Visit})=1-0.1\cdot P(\text{Eggs|Visit}).$$
I have to find $$P(\text{Eggs|Visit})=\frac{e^{-1}(-1)^0}{0!}=\frac{1}{e}\approx 0.3679.$$
So
$$P(\text{No eggs})=1-0.1\cdot0.3679=0.9632.$$
Correct answer is $0.9 + 0.1\cdot e^{-1}\approx0.94.$
Questions:
1) In a problem, what is the main indicator that Baye's theorem should be used?
2) I don't really understand how it suffices to directly use the pmf for Possion distribution to compute $P(\text{No eggs}).$ My random variable $X$ only counts ne number of eggs on a leaf, not number of eggs given a visitor. I think this has something to do with getting incorrect answer.
Best Answer
What you have computed as $1/e$ is $P[\text{No eggs|Visit}]$.
The probability of no eggs is: Probability of no visit + Probability[visit] $\times$ (Probability of no eggs given visit).