Answer:
The expectation is [tex]E(1 )= -\$ 1[/tex]
Step-by-step explanation:
From the question we are told that
The first offer is [tex]x_1 = \$ 8000[/tex]
The second offer is [tex]x_2 = \$ 4000[/tex]
The third offer is [tex]\$ 1600[/tex]
The number of tickets is [tex]n = 5000[/tex]
The price of each ticket is [tex]p= \$ 5[/tex]
Generally expectation is mathematically represented as
[tex]E(x)=\sum x * P(X = x )[/tex]
[tex]P(X = x_1 ) = \frac{1}{5000}[/tex] given that they just offer one
[tex]P(X = x_1 ) = 0.0002[/tex]
Now
[tex]P(X = x_2 ) = \frac{1}{5000}[/tex] given that they just offer one
[tex]P(X = x_2 ) = 0.0002[/tex]
Now
[tex]P(X = x_3 ) = \frac{5}{5000}[/tex] given that they offer five
[tex]P(X = x_3 ) = 0.001[/tex]
Hence the expectation is evaluated as
[tex]E(x)=8000 * 0.0002 + 4000 * 0.0002 + 1600 * 0.001[/tex]
[tex]E(x)=\$ 4[/tex]
Now given that the price for a ticket is [tex]\$ 5[/tex]
The actual expectation when price of ticket has been removed is
[tex]E(1 )= 4- 5[/tex]
[tex]E(1 )= -\$ 1[/tex]
