# binomial random variables :-/

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• Oct 9th 2007, 08:44 AM
inneedofhelp
binomial random variables :-/
A scalper is considering buying tickets for a particular game. The price of the tickets is \$75, and the scalper will sell them at \$150. However, if she can't sell them at \$150, she won't sell them at all. Given that the demand for tickets is a binomial random variable with parameters n = 10 and p = 1/2, how many tickets should she buy in order to maximize her expected profit?
• Oct 9th 2007, 08:53 AM
TKHunny
Just walk your way through it.
Profit From Sale: 150 - 75 = 75
Profit From No Sale: 0 - 75 = -75

Buy 1
Pr(Not Selling It) = ½^10 = q
Pr(Selling It) = 1 - q = p
Expected Profit: p*(75) + q*(-75)

Buy 2
Pr(Selling Zero) = ½^10 = q
Pr(Selling 1) = 10*½^10 = w
Pr(Selling Both) = 1 - (q+w) = p
Expected Profit: p*(150) + w*(0) + q*(-150)

You can keep wandering down the entire Binomial distribution (a nice spreadsheet makes this relatively simple) or you can determine a more direct way if you are REALLY paying attention.

Note: It is possible the right answer if greater than 10. I didn't really solve the problem, but I'd keep that in mind, just in case you need it.