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I have a tool that, every time I use it, it has a 1% chance of breaking. What is the expected number of uses of this tool, before it breaks $?$.

I think that the probability of the tool breaking on the $n^{\rm th}$ use is $$ \operatorname{P}\left(n\right) = {1 - \operatorname{P}\left(n - 1\right) \over 100}\quad\mbox{with}\quad \operatorname{P}\left(1\right) = {1 \over 100} $$ and I guess I need to sum over all $n \times \operatorname{P}\left(n\right)$, but is there a way to find an exact answer to the question $?$.

Felix Marin
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Ricardo Fodra
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1 Answers1

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The intuitive answer is that it is 100. What else could it be.

But, since intuition has led you astray in the past....

What is the chance that the tool breaks on the n'th use. That means that it didn't break for n-1 uses. And broke once.

$P(x=n) = (1-0.01)^{n-1}(0.01)$

$E[X] = \sum_\limits{n=1}^\infty nP(x=n) = 0.01\sum_\limits{n=1}^\infty n0.99^{n-1}$

There is a formula for that summation.

$\sum_\limits{n=1}^\infty nx^{n-1} = \frac {d}{dx}\sum_\limits{n=1}^\infty x^{n} = \frac {d}{dx} \frac {x}{1-x} = \frac {1}{(1-x)^2}$

I hope you don't mind my use of calculus here.

$E[X] = \frac {0.01}{(1-0.99)^2} = 100$

Just like your intuition suggested!

user317176
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