Binomial Probability, finding the number of trials.

Question

Question:

For the above question, the probability of an adult having a slow heart rate is $0.1587$, so the the success of choosing an adult with a slow heart rate is, $p = 0.1587$.

Here is what I have tried with the question:

Since $n$ has to be integral value (we are talking about people). We can say that Pr$(P_n > \frac{1}{n})$ is the same as saying Pr$(P_n > 1)$. So if Pr$(_n > 1) > 0.99$, then $1 - (\text{Pr}(P_n = 0) + \text{Pr}(P_n = 1)) > 0.99$. This can be then simplfied to $\text{Pr}(P_n = 0) + \text{Pr}(P_n = 1) < 0.01$. How would I continue from here to find the least number of trials, n for this question?

Answer 1

$p = 0.1587$, where $p$ is the probability of a slow heart rate. then $(1-p) = 0.8413$

So you need to find $n$ such that,

$ \displaystyle {n \choose 0} (1-p)^n + {n \choose 1} (1-p)^{n-1} ~ p \lt 0.01$

$0.8413^n + n \times 0.8413^{n-1} \times 0.1587 \lt 0.01 $

You need an online tool or a small program to solve for $n$. Using WolframAlpha, $n \geq 39$.