Probability of success in $n$ trials

Question

I'm stuck on my statistics homework and would appreciate your help.

Question: Repeated independent trials of a certain experiment are carried out. On each trial the probability of success is $0.12$. Find the smallest value of $n$ such that the probability of at least one success in $n$ trials is more than $0.95$.

My attempt:

$X \sim B(n,0.12)$

$P(X \ge 1) = 1-P(X \le 1)$

$P(X \le 1) = 0.88^n + 0.12n(0.88)^{n-1}$

$0.88^n + 0.12n(0.88)^{n-1} \gt 0.05$

After a page of algebra I reached an equation that I couldn't solve (of the form $e^x +bx + c = 0$) so I knew I must be doing something wrong. Is this the correct way to approach the problem? How else can I do it? What am I doing wrong?

Answer 1

Oops, you made a mistake. $P(X\geq1)=1-P(X=0)$. You have it as $P(X\geq1)=1-P(X=0)-P(X=1)$

Answer 2

Hint: $P(X \geq 1)=1-P(X=0)=1-0.88^n\geq 0.95$

Answer 3

$P($at least one success$) + P($no success$)=1 \implies P($at least one success$)=1-(0.88)^n$. I think you can proceed from here.