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In a 12 hour period, there can be 0, 1, or 2 new mutations with probabilities 0.8, 0.15, 0.05 respectively. Suppose that testing stops as soon as a cell becomes malignant. What is the expected number of mutations that have occurred at this time? The no of tests to malignancy is geometric with parameter $p = 1/20$

My thoughts:

Let $E[X_i] = 0.8*0 + 0.15*1 + 0.05*2 = 0.25$

I have a feeling I can use Wald's equation here, but am not sure how to find my $E[N]$

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Yes.   Good decision to use Wald's Lemma; that $~\mathsf E(\sum_{k=1}^N X_k) = \mathsf E(N)\,\mathsf E(X_1)~$ when the sequence of $(X_k)$ are independent and identically distributed random variables (and each independent of $N$ too).

Also, $\mathsf E(X_1)=0.25$ as you calculated.

So, you have been told that the count of tests before malignancy is geometrically distributed with parameter $p=\tfrac 1{20}$.   That count is what $N$ represents.

Now, what is the expectation of a geometric random variable?

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    So in our case, $E[N] = 1/p = 20$? So essentially the expected number of mutations that occur when a cell becomes malignant is $20*0.25$? – 2017-01-27