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This is another one of the questions I didn't get a chance to ask the TA at my school today. I thought I had a pretty good grasp on Expectations but apparently I could still use some clarification. Hopefully someone can help me.

Suppose that $n$ people take a blood test for a disease, where each person has probability $p$ of having the disease, independent of other persons. To save time and money, blood samples from $k$ people are pooled together. If none of the $k$ persons has the disease then the test will be negative, but otherwise it will be positive. If the pooled test i sportive then each of the $k$ persons is tested separately (so $k+1$ tests are done in that case)

Let $X$ be the number of tests required for a group of $k$ people. Show that

$$ E(X)=k+1-k(1-p)^{k} $$

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up vote 2 down vote accepted

There are two possibilities:

(i) Nobody has the disease. This has probability $(1-p)^k$;

(ii) At least $1$ person has the disease. This has probability $1-(1-p)^k$.

In Case (1), the cost is $1$. In Case (ii), the cost is $1+k$. So if $X$ is the cost, then $$E(X)=(1-p)^k(1) +\left[1-(1-p)^k\right](1+k).$$ We can simplify the above expression a little.

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That makes a lot more sense now! It didn't really occur to me that I needed to consider each of the k people being diagnosed as separate cases. I just considered it as a whole. Thank you for the clarification :) – TopGunCpp Nov 13 '12 at 5:36

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