Expectation - June 19

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Expectation - June 19

Amit Goyal
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AJ
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Re: Expectation - June 19

AJ
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! :(
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Re: Expectation - June 19

Vasudha
how about v say that the expected no of tests= no of groups*expected no of tests per group? that would be (n/k)[[(1-p)^k+(k+1)(1-(1-p)^k)]?
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Re: Expectation - June 19

Amit Goyal
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(N/k) + N [1 - (1 - p)^k]