# Expectation - June 19 Classic List Threaded 4 messages Open this post in threaded view
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## Expectation - June 19

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## Re: Expectation - June 19

 Number of groups= N/k Probability that a particular group is tested negative= Probability that each person in the group had a negative test = (1-p)^k = Say, x (for convenience in writing) Probability that a particular group is tested positive= 1-(1-p)^k = 1-x Now using binomial: No. of trials= N/k Success= A group tested negative failure= A group tested negative Expectation(No. of successes)= N/k (x)(1-x) = Expected no. of groups tested negative. Now, when test for group is negative: No. of tests required=1         when test for group is positive: No. of tests required= K+1 Expected no. of tests= 1*[N/k (x) (1-x)] + [k+1]*[N/k - N/k (x) (1-x)]                              = [N/k (x) (1-x)] + [k+1]N/k - [k+1][N/k (x) (1-x)]                              = [N/k (x) (1-x)] [1-k-1] + [k+1]N/k                              = -N(x)(1-x) + [k+1] N/k                              = N[1 + 1/k - ((1-p)^k)(1-(1-p)^k)] Is it correct? It seemed fine until I came to the final answer... But now it looks wrong! :(