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There is a $30%$ chance of a car having a leaky tank. The probability an inspector will have to check at least $n$ cars to find the first one with a leaky tank is $0.05$. Find $n$.

Let $X = \text{The number of non leaky cars before first leaky}$

Then we want $P(X \ge n)$ is this correct?

The function $f(x) = P(X = x) = (0.7)^x 0.3$ correct?

Thus we want $P(X \ge n) = f(n) + f(n+1) + f(n+2) + ... $

But the book insists on it being $P(X \ge n - 1)$

Why $n-1$?

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    No, he has to check $n$ _cars_ and the last one will have a leaky tank, so he will have checked $n-1$ _cars without a leaky tank_ and the last car with a leaky tank. – 2017-02-09
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    "...n cars to find the first..." means "n cars and the n-th is the first to leak", unlike "...n cars **before** finding the first leaky car" – 2017-02-09

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