Answer:
Suppose that in a given event, there are N outcomes:
{x₁, x₂, ..., xₙ}
Such that the probabilities are:
{p₁, p₂, ..., pₙ}
The expected value is:
EV = x₁*p₁ + x₂*p₂ + ... + xₙ*pₙ
In this case, we know that each ticket costs $3.
There is 1 winning ticket out of 300 tickets, where the prize is $62.
Then we have two events, the one where you win.
x₁ = $62 - $3 = $59
With a probability:
p₁ = 1/300
The one where you don't win (and still pay for the ticket):
x₂ = -$3
p₂ = 299/300
a) Then the expected value is:
EV = ( 1/300)*$59 + (299/300)*(-$3) = -$2.79
b) The expected value is additive, this means that if you buy 6 tickets, the expected value will be equal to 6 times the expected value of a single ticket.
EV = 6*(-$2.79) = -$16.74
c) For the PTO the events are:
Winner ticket, so you lose $62 and win $3 for selling the ticket.
x₁ = $3 - $62 = -$59
p₁ = 1/300
Normal ticket, you win $3.
x₂ = $3
p₂ = 299/300
In this case the expected value is:
EV = (1/300)*(-$59) + (299/300)*$3 = $2.79
d) If the PTO sells the 300 tickets, then the expected value is 300 times the expected value for a single ticket:
EV = 300*$2.79 = $837
We should have the exact same than selling the 300 tickets:
300*$3 = $900
minus the $62 for the price:
$900 - $62 = $838
There is a difference of $1, that happens because we rounded the value of the expected value (the actual one was $2.792...)
