# Improper Use of Geometric Formula

Assume that every time you hear a song on the radio, the chance of it being your favorite song is $$2\%$$. How many songs must you listen to so that the probability of hearing your favorite song exceeds $$90\%$$?

My initial approach was:

This is a geometric distribution with probability of success $$p=0.02$$. Let the random variable $$X$$ be the number of songs heard BEFORE I hear my favorite song. For example, $$X=3$$ means I heard 3 mediocre songs before my favorite song. So we want,

$$P(X=k)=(1-p)^kp > 0.9\\\\ ~~~~~~~~~~~~~~~~\Rightarrow (1-p)^k > 0.9/p\\\\ ~~~~~~~~~~~~~~~~\Rightarrow k(\log(1-p)) > \log(0.9/p)\\\\ ~~~~~~~~~~~~~~~~\Rightarrow k > \frac{\log(0.9/p)}{\log(1-p)}\\\\ ~~~~~~~~~~~~~~~~= k > \frac{\log(0.9/0.02)}{\log(0.98)}\\$$

The correct approach was:

$$P$$(good song) $$=0.02$$

$$P$$(bad song) = $$0.98$$

$$P$$(n bad songs) = $$0.98^n$$

$$P$$(good song after n) = $$1-(0.98)^n$$

thus,

$$1-(0.98)^n > 0.9 \Rightarrow n > \frac{\log{(1-0.9)}}{\log{0.98}}$$

What did I do wrong in my initial approach?

You are considering the probability that you hear exactly $$k$$ songs before the favourite, which is not greater than $$2\%$$ for all $$k$$, let alone $$90\%$$. This is illustrated if you evaluate the final expression for $$k$$ in your approach – it comes to the absurd $$-188.423$$.
You want the value of $$n$$ such that the probability for hearing your favourite song among $$n$$ songs exceeds $$90\%$$.   Not only "at the end", but anywhere "among" them.
Alternatively: that the probability for not hearing your favourite song among those $$n$$ is at most $$10\%$$.
$$(1-0.02)^n\leqslant (1-0.90)\\ n\log 0.98\leqslant \log 0.10\\\vdots\\ n\geqslant 114$$