# Thread: Data Management - Prob. Distributions

1. ## Data Management - Prob. Distributions

A lottery ticket has a $1,000,000 first price,a$25,000 second, and 5 $1000 third prizes. a total of 2,000,000 tickets are sold. If a ticket costs$2.00, what is the expected profit per ticket ?

I tried E(x) =Sum Xi . P(xi)

I am not getting 1.48 as an answer...

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2. 2,000,000 tickets will sell for $4,000,000. Now the lottery company gives away one$ 1,000,000 award, one $25,000 award and five$ 1,000 awards.

Summing these awards up, you have $1,030,000 in awards given away. So the lottery company expects to make a total of$ 4,000,000 - $1,030,000 =$ 2,970,000.

There are 2,000,000 tickets, so the expected profit per ticket is $\displaystyle \frac{ \$2,970,000}{2,000,000}$3. Why did you add ? Isnt the expectation E(X) = N x P ? 4. Doesn't really matter. Letting X = the amount a ticket buyer wins,$\displaystyle E(X) = \$1,000,000 \times \frac{1}{2,000,000} +$$\displaystyle \$ 25,000 \times \frac{1}{2,000,000} + \$1,000 \times \frac{5}{2,000,000} = \$ 0.515$Now, if the ticket buyer payed$2 and can expect to win $0.515, this means the lottery company expects to make the rest from every ticket sold (i.e.$2 - $0.515 =$1.485)

This gives the same result as my first post.

5. Originally Posted by Faisal2007
Isnt the expectation E(X) = N x P ?
What's P meant to be? It looks to me like you're talking about the formula for the expected value of a random variable that follows a binomial distribution. The binomial distribution is not releveant here.

You need to review your class notes and/or textbook and get things straight.

I will add that the formula for the expected value of a binomial random variable is derived using the general definition of expected value:

$\displaystyle E(X) = \sum_{i = 1}^{\infty} x_i \cdot \Pr(X = x_i)$.