Given:
The number of tickets = 1000
Mathew buys a single ticket for $8
Winning price = $800
To find:
The expected earning per ticket.
Solution:
It is given that the total number of tickets is 1000. So, the probability of winning is:
[tex]P(Winning)=\dfrac{1}{1000}[/tex]
The probability of losing is:
[tex]P(losing)=\dfrac{999}{1000}[/tex]
The expected earning per ticket is:
[tex]E(x)=800\times P(Winning)-8\times P(losing)[/tex]
[tex]E(x)=800\times \dfrac{1}{1000}-8\times \dfrac{999}{1000}[/tex]
[tex]E(x)=\dfrac{800}{1000}-\dfrac{7992}{1000}[/tex]
[tex]E(x)=\dfrac{800-7992}{1000}[/tex]
[tex]E(x)=\dfrac{-7192}{1000}[/tex]
[tex]E(x)=-7.192[/tex]
Therefore, the expected earning per ticket is -7.192. Here, negative sign means loss of $7.192.
