Answer:
EV = -$3.50
Price = $1.50
Step-by-step explanation:
By purchasing one ticket, a person has a 1 in 1000 probability of winning the laptop computer worth $1,500 (with a value of $1,495 discounting the ticket price) and a 999 in 1000 probability of losing $5. The expected value is:
[tex]EV = \frac{1}{1000}*1495 -(\frac{999}{1000}*5)\\EV = -\$3.50[/tex]
The expected value of the purchase of one ticket is -$3.50.
In order for the club to not lose any money, the total cost of the 1000 tickets must equal the value of the computer:
[tex]P = \frac{\$1,500}{1,000}\\P=\$1.50[/tex]
The smallest purchase price for which the club will not lose any money IS $1.50
