Thread: discrete random variable

Are you sure that one option isn't $36.40 and not $86.40?.

Remember, you're buying 5 tickets, not 3.

Generally in a raffle you do not get your money for the tickets back if you win. So there is 100% chance of losing the $100.

i cant believe i was thinking of 3 tickets. ya 5
and the answers could be wrong, the problems are known to have errors

Originally Posted by cubs3205

A local teacher's education association decides to hold a raffle. They are going to sell 500 tickets for $20.00 each. 1st Prize is a $5000 scholarship, 2nd Prize is a $1000 scholarship, and the 3rd Prize is a $500 scholarship. You decide to purchase $100 worth of tickets. What are your expected winnings?
a -17.28
b -86.40
c -0.80
d 20.00

I approached it as this
there is 3/500 chance of winning $5000
there is 3/500 chance of winning $1000
there is 3/500 chance of winning $500
and t here is 497/500 chance of losing the 100
so i summed the probabilities and winnings
(3/500)4900 + (3/500)(900) + (3/500)(100) + 497/500(-100)
i don't get any of the answer choices
what am i doing wrong?

To do this properly you should construct a contingency tree of the
possible outcomes:

(win 1st, not win 2nd, not win 3rd),
(win 1st, win 2nd, not win 3rd),
... etc

Then you would multiply the terminal leaf probabilities by the prizes
corresponding to the leaf probabilities and sum to get the total
expected prize money, then subtract your stake to get your expected
winnings.

However this is a multiple choice question, so we can do an
approximate calculation and choose the closest answer to this.

You have 1/100 of the tickets, so you should expect to take away
something like 1/100 of the prize money, or $65, at a cost of $100.
So we should expect to see a return of about -$35, but this is not
close to any of the listed answers!?

In retrospect I suspect there is nothing approximate about this argument,
but I can't be bothered to prove it here.

RonL