Poisson distribution independent events

Question

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: $$\sum_{i=1}^{\infty}P(i)\times (1-0.2)=0.8\times 0.95021.....$$

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

Is this correct?

Answer 1

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^i\frac{\lambda^i}{i!}\exp(-\lambda)\\ &=\exp(-0.2\lambda)\sum_{i=0}^\infty \frac{(0.8\lambda)^i}{i!}\exp(-0.8\lambda)\\ &=\exp(-0.2\lambda)\\ &=\exp(-0.6)\end{align}

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

Answer 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 $$ \sum_{i=0}^\infty P(i) \cdot (0.8)^i. $$

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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.