Conditional Probability and Shark Attacks

Question

Probability of being attacked on day n by shark given that you have not being attacked on days before n is $\cfrac{1}{n + 1}$. What is the probability of the number of the day when shark attacks for the first time is n.

Is it ok to think in the following way:

$\mathbb{P}($day of first attack = n$) = \mathbb{P}($attack on day n $\land$ $\neg$attack on day $(n-1)$ $\land$ $...$ $\land$ $\neg$ attack on day 1$)$ $= \mathbb{P}($attack on day n $|$ $\neg$attack on day $(n-1)$ $\land$ $...$ $\land$ $\neg$ attack on day 1 $) \cdot \mathbb{P}($ $\neg$attack on day $(n-1)$ $\land$ $...$ $\land$ $\neg$ attack on day 1 $)$ $=\cfrac{1}{n + 1} \cdot \Big( 1 - \cfrac{1}{n} \Big)\cdot ... \cdot \cfrac{1}{2} = \cfrac{1}{n + 1} \cdot \cfrac{n - 1}{n} \cdot \cfrac{n}{n - 1} \cdot ... \cdot \cfrac{1}{2} = \cfrac{1}{(n + 1)\cdot n}$

Answer 1

That is correct; for completeness you might want to add your derivation of the final answer (cancelling stuff from fractions). But otherwise, that's correct!

Edit response to comment

Yup, that's correct! I was actually going to point that out as an interesting side effect of the conclusion, but by the time I noticed I'd already posted the reply and didn't want to self-comment or edit the post. But that is correct, because the sum of the probabilities that you get attacked on days $n=1,2,3,\ldots$ is $1$ so you will always get attacked at some point.

Answer 2

Yes.   The probability for the first attack being on day $n$ is clearly the probability for being attacked on that day and not attacked on any of the prior days, and you have calculated that well.

$$\begin{align}\mathsf P(N=n)&=\dfrac{1}{n+1}\prod\limits_{k=1}^{n-1}\dfrac{k}{k+1}\\[1ex]&=\dfrac 1{n+1}~\dfrac{n-1}{n}\dfrac{n-2}{n-1}\cdots\dfrac{1}{2}\\[1ex]&=\dfrac{1}{n~(n+1)}\end{align}$$