A mail order company offers their first $1000$ customers a ladies' or mens' watch. Suppose that both sexes are equally attracted by the offer. How many ladies' and mens' watches are needed in order to ensure that –with a probability of at least $98$%, all $1000$ customers receive a matching watch.
I was tipped to use Chebyshev's inequality.
My idea:
Define $X:=$"Number of male watches stocked"
Note that since "both sexes are equally attracted by the offer", I only need to look at $500$ watches.
My problem is that my definition of RV $X$ does not allow for me to obtain $\mathbb E[X]$ let alone $\operatorname{Var}(X)$. I believe that I am using the wrong random variables, but cannot find any other appropriate ones.
Best Answer
Let $X=500+x$ male watches stocked, and the same number of female watches stocked. Note that this is not a random variable. Here the number of men or women is random since any customer can be of any sex equiprobable. Denote by $S_{1000}$ the number of men among first $1000$ customers. As wueb suggests, $S_{1000}=X_1+\ldots+X_{1000}$ where $X_i$ equals to $1$ if $i$-th customer is a men and $X_i=0$ if it is a women. Then $1000-S_{1000}$ is the total number of women.
We need both $S_{1000}\leq 500+x$ and $1000-S_{1000}\leq 500+x$ with probabilility at least $0.98$. These inequalities transform to $$ 500-x\leq S_{1000}\leq 500+x $$ or $$ |S_{1000}-500|\leq x. $$ We need to find $x$ such that $$ \mathbb P\left(|S_{1000}-500|\leq x \right) \geq 0.98. $$ If it is required to use Chebyshev's inequality, use $\textrm{Var}(S_{1000})=1000\cdot \textrm{Var}(X_1)=250$ in r.h.s.
$$ \mathbb P\left(|S_{1000}-500|\leq x \right) = 1- \mathbb P\left(|S_{1000}-500|> x \right)=1- \mathbb P\left(|S_{1000}-500|\geq x-1 \right) $$ $$ \geq 1-\frac{\textrm{Var}(S_{1000})}{(x-1)^2}\geq 0.98. $$ Find minimal integer $x$ satisfying this inequality and $500+x$ will be the answer for number of male watches.