I'm not sure if I'm supposed to use a Poisson distribution or Conditional Probability (or both) to answer this question

Question

I have a question that I'm trying to solve. I have the answer but I don't know how they arrived at the answer so I can't compare my work and see where I went wrong.

The number of injury claims per month is modeled by a random variable N with

P(N=n) = 1/(n+1)(n+2), n>= 0

Determine the probability of at least one claim during a particular month, given that there have been at most four claims during the month.

To me this seemed like I had to use conditional probability instead of a poison distribution. My work was as follows:

P(N=0) = 1/2
P(N=1) = 1/6
P(N=2) = 1/12
P(N=3) = 1/20
P(N=4) = 1/30

I tried to calculate the condition probability

P(N=1 | P(N=1 or N=2 or N=3 or N=4).

I got

P(N=1 or N=2 or N=3 or N=4) = (1/6) + (1/12) + (1/20) + (1/30) = .33

and I used these numbers to calculate

P(N=1)P(N=1 or N=2 or N=3 or N=4)/P(N=1 or N=2 or N=3 or N=4)

The answer that I get is .167 but my answer key says that the correct answer is .4. I'm obviously way off. Any guidance would be appreciated.

Answer 1

If there have been four claims at most, you need to consider values of $N$ from $\color{blue}{0}$ to $4$ inclusive, hence the total probability is $$P(N=0)+P(N=1)+P(N=2)+P(N=3)+P(N=4)=\frac{5}{6}$$ For the probability of at least one outcome, the values of $N$ from $1$ to $4$ need to be considered $$P(N=1)+P(N=2)+P(N=3)+P(N=4)=\frac{5}{6}-P(N=0)=\frac{1}{3}$$ Thus the probability of at least one claim given that there are have been at most $4$ claims is $$P(N\geq1|N\leq4)=\frac{P(N=1)+P(N=2)+P(N=3)+P(N=4)}{P(N=0)+P(N=1)+P(N=2)+P(N=3)+P(N=4)}\\=\frac{1}{3}\times\frac{6}{5}=\frac{2}{5}=0.4$$

Answer 2

I would say

Event A: "at least one" --> N = 1 or 2 or 3 or 4 --> P(A) = 1/3

Event B: "at most four claims" --> N = 0 or 1 or 2 or 3 or 4 --> P(B) = 5/6

$P(A|B) \cdot P(B) =P(A \cdot B) =P(B|A) \cdot P(A) \Rightarrow P(A|B) \cdot P(B)=P(B|A) \cdot P(A) \Rightarrow P(A|B)=\frac{P(B|A) \cdot P(A)}{P(B)}$

Because $P(B|A)=1,\,\,$ then $\displaystyle P(A|B)=\frac{1 \cdot \frac{1}{3}}{\frac{5}{6}}=0.4$