SOLUTION: Could you please help me with this? Thank you 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 p

Algebra ->  Proofs -> SOLUTION: Could you please help me with this? Thank you 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 p      Log On


   

Question 145212This question is from textbook
: Could you please help me with this? Thank you
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 jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a) You are correct.

b) To find the fair price, replace the 2 with x and the -1.14 with 0 (since a "fair" price means that the player will break even) to get

400%281%2F1000%29+%2B+200%281%2F1000%29+%2B+50%285%2F1000%29+-x%28993%2F1000%29=0


4%2F10+%2B+2%2F10+%2B+1%2F4+-x%28993%2F1000%29=0 Multiply


17%2F20+-x%28993%2F1000%29=0 Combine like terms


+-x%28993%2F1000%29=-17%2F20 Subtract 17%2F20 from both sides


Multiply both sides by -1000%2F993 to completely isolate x



+x=850%2F993 Multiply


+x=0.8559919 Use a calculator to approximate the fraction


x=0.86 Round to the nearest hundredth


So the fair price of the ticket is about 86 cents.