My assignment is to find the number of trials in order to get a probability greater of 95% given that each success is 25%. My first thought was to use Binomial Distribution: $$ \ P(X>=n) = 1 - nC0*p^(0)(1-p)^(n) \\ \ .95 >= 1 - (1-.25)^(n) \\ \ 1-.95 <=(1-.25)^(n) \\ \ n >= log(1-.95)/log(1-.25) $$
I got that n >= 11 trials are needed to get a percentage greater than 95%
If I understand correctly, you have a random variable $X \sim B(n, 0.25)$ and you want to find the smallest value of $n$ such that $P(X \geq 1) \geq 0.95$.
Because the binomial distribution is descrete, $P(X \geq 1) = 1 - P (X < 1) = 1 - P(X = 0)$. So you're looking for the smallest value of $n$ such that $1 - P(X = 0) \geq 0.95$.
For any binomially distributed variable, $P(X = 0) = (1 - p) ^ n$.
So this gives you $1 - (1 - (0.25)) ^ n \geq 0.95$. Solve for n and round up to the nearest integer (giving 11, so you were correct).