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Is it possible to calculate the mean of a Poisson Distribution, given the probability of a single event occurring?

Specifically, this is the question. Apples in a particular farm contain some amount of metal, which follows the normal distribution $N(16,16)$. An apple is considered to be acceptable if the amount of metal is under 18 units. Apples are randomly selected and tested for metal one by one. Assuming the metal test is absolutely accurate, find the probability that more than two tests are required for obtaining the first acceptable apple.

I've done some normal distribution calculations to obtain the probability that an apple is acceptable, which is $P(X)=0.6915$. Now, how can I find the probability that more than two tests are required to obtain the first acceptable apple? I thought of using a Poisson distribution like $$P(W \gt 2)=1-P(W=0)-P(W=1)-P(W=2)$$ But to calculate $P(X=x)$, I would need to know the mean. So how do I calculate the mean?

Unless we could use some other method to answer the question... I'm not good at statistics so using Poisson is the only thing I could think of using. (I'm still interested in knowing whether we can calculate mean from probability, though.)

Thanks!

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    $\begingroup$ Indeed, one wonders how Poisson distributions came to even be mentioned at all. $\endgroup$ – Did Dec 1 '16 at 12:14
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"[...] the probability that more than two tests are required for obtaining the first acceptable apple"

The opposite of this event is much easier to work with: {trial 1 is acceptable} or {trial one is not acceptable, but trial 2 is}. Then $$ P(\text{trial acceptable}) = P(X\leq 18) \stackrel {\text{def}}=p \approx 0.6915. $$ These are bernoulli trials (not Poisson) and we get $$ P(\{\text{trial 1 acceptable}\} \text{ or } \{\text{trial 1 not acceptable, trial 2 acceptable}\}) = p+(1-p)p. $$ The answer is then $$ P(\text{more than two trials required before success}) = 1-(p+(1-p)p) \approx 0.095. $$

I assume that by $X$ you mean $X\sim \mathcal N(16,16)$

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