SOLUTION: not sure if this is right 6. (4 pts) 1000 tickets for prizes are sold for $2 each. Seven prizes will be awarded – one for $400, one for $200, and five for $50. Steven purchas

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Geometry -> SOLUTION: not sure if this is right 6. (4 pts) 1000 tickets for prizes are sold for $2 each. Seven prizes will be awarded – one for $400, one for $200, and five for $50. Steven purchas      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 144879This question is from textbook
: not sure if this is right
6. (4 pts)
1000 tickets for prizes are sold for $2 each. Seven prizes will be awarded – one for $400, one for $200, and five for $50. Steven purchases one of the tickets.
400(1/1000) + 200(1/1000) + 50(5/1000) – 2(993/1000) = -1.14
a) Find the expected value expected value is -$1.14
b) Find the fair price of the ticket. ?????
This question is from textbook

Answer by solver91311(6137) About Me  (Show Source):
You can put this solution on YOUR website!
You have part a done correctly. Steven (or anyone else for that matter) can expect to lose -$1.14 for every $2.00 spent. In other words, the game is rigged (way, way, way) in favor of the house.

The fair price for a ticket is that price that would make the expected value be zero. So, reuse your expected value formula, this time using a variable P%5Bf%5D in place of the $2 in your previous calculation and set it equal to 0:

400%281%2F1000%29+%2B+200%281%2F1000%29+%2B+50%285%2F1000%29+-%28P%5Bf%5D%29+%28993%2F1000%29+=+0

Solve for P%5Bf%5D and you have your fair price.