Dtjytyk

I am going to write my solution upon an earlier suggestion made on this Poisson distribution problem, I would appreciate if someone could tell me if it is correct:

Number of physics problems that Mike tries for any given week follows a Poisson distribution with $μ=3$.

Every problem that mike tries is independent of one another, and has a constant probability of $0.2$ of getting the problem correct. (Mike's number of tries at the problems is independent of him answering a problem correctly).

Question: What is the probability that mike answers no questions correctly in any of the given two weeks?

My Solution: So if we find the probability $P(i)$ where it stands for the probability of attempts at $i$ problems, then we simply can calculate from the Poisson cdf, of mean $mu=3$, hence we have:

$$sum_{i=1}^{infty}P(i)=text{PoissonCdf}(X>0,mu=3)=0.95021.....$$
This gives the total probability of all possible attempts made at the problem. The probability of getting all these attempts wrong will be:
$$sum_{i=1}^{infty}P(i)times (1-0.2)=0.8times 0.95021.....$$

However it asks for two weeks, and it is independednt hence we multpiply this value by itself:
$$bigg(sum_{i=1}^{infty}P(i)times (1-0.2)bigg)^2 =(0.8times 0.95021.....)^2$$

Is this correct?

Let $N$ be the number of question that he answers in a week.

The probability that in a week, he answer no answer correctly is

begin{align}sum_{i=0}^infty P(X=0|N=i)P(N=i)&=sum_{i=0}^infty (1-0.2)^iP(N=i)\
&= sum_{i=0}^infty 0.8^ifrac{lambda^i}{i!}exp(-lambda)\
&=exp(-0.2lambda)sum_{i=0}^infty frac{(0.8lambda)^i}{i!}exp(-0.8lambda)\
&=exp(-0.2lambda)\
&=exp(-0.6)end{align}

Hence the answer is $exp(-1.2)$.

Your answer isn't quite right. It's fine to look at $P(i)$, the probability of attempting $i$ problems, but then he must get all $i$ incorrect, so multiply by $(0.8)^i$ and sum over $i$. (You are multiplying every term by $0.2$.)
Note that attempting $0$ questions is also fine, as then he certainly doesn't answer any correctly!

So the answer for one week is
$$ sum_{i=0}^infty P(i) cdot (0.8)^i. $$

When you multiply $P(i) = e^{-mu} mu^i / i!$ by $(0.8)^i$, what do you notice? It's pretty similar to a Poisson cdf!
In fact, this is called "Poisson thinning", which roughly says that if you have a Poisson process of rate $lambda$ and accept/reject arrivals with probability $p$, then you get a Poisson process of rate $lambda p$.
The answers form a Poisson process of rate $3$ and are right with probability $0.2$, so the correct answers form a Poisson process of rate $3 cdot 0.2 = 0.6$.
The answer then is
$$ P( text{Poisson}(0.6) = 0 ) = exp(-0.6). $$
You can read more about this in the my supervisor's lecture notes---see specifically Section 1.4.