Suppose that the ratio of males to females in the house is x, i.e. there are x males for each female. Then it follows that the probability that a randomly chosen toilet visit is made by a male equals x/(x+1) and that it is made by a female equals 1/(x+1). If we denote the total number of toilet visits by k, then the expected number of male visits among them is just the product of k and x/(x+1)which is kx/(x+1). Under Scheme 1, since 2 units of work are done during a male visit and none at all during a female visit, the expected total amount of work done in k visits is 2kx/(x+1).

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