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) = $1-(0.98)^n$


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

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 $10\%$.


$$(1-0.02)^n\leqslant (1-0.90)\\ n\log 0.98\leqslant \log 0.10\\\vdots\\ n\geqslant 114$$


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