
Number of groups= N/k
Probability that a particular group is tested negative= Probability that each person in the group had a negative test = (1p)^k = Say, x (for convenience in writing)
Probability that a particular group is tested positive= 1(1p)^k = 1x
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)(1x) = 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) (1x)] + [k+1]*[N/k  N/k (x) (1x)]
= [N/k (x) (1x)] + [k+1]N/k  [k+1][N/k (x) (1x)]
= [N/k (x) (1x)] [1k1] + [k+1]N/k
= N(x)(1x) + [k+1] N/k
= N[1 + 1/k  ((1p)^k)(1(1p)^k)]
Is it correct? It seemed fine until I came to the final answer... But now it looks wrong! :(
