Expected Value Exercise 2
John Smith is considering the purchase of a used car
that has a bank book value of $16,000.
He believes that there is a 20% chance that the car's transmission is
damaged. If the transmission is damaged,
the car would be worth only $12,000 to Smith.
What is the expected value of the car to Smith?
EXPECTED VALUE = Probability x Payoff of all possible outcomes
ReplyDeleteEU = (0.8 x 16000) + (0.2 x 12000) = 15.200
Known:
ReplyDeletePrice if the cars is not damaged : $ 16000
Price if the cars is damaged : $ 12000
Chance that the car’s transmission will be broke : 20% = 20/100 = 0.2
Chance that the car’s transmission will be in good condition : 100% - 20% = 80% = 80/100 = 0.8
Solution:
Expected Value = (Price undamage) (Probability undamage) + (Price damage) (Probability damage)
= ( 16000 x 0.8 )+( 12000 x 0.2 )
= 12800 + 2400
= $ 15200
So, the EXPECTED VALUE of the car is $15200
EU= pr(damaged)(the price of damaged car)+pr(not damaged)(the price of not damaged car)
ReplyDeleteEU= (80/100x16,000)+(20/100x12,000)
EU= 15,200
E(U)=(Prob. of Damaged Transmission)x(Exp. Val. of Damaged Car)+(Prob. of No Damage To Transmission(of 1 - 20%)x(Exp. Val. of Undamaged Car)
ReplyDeleteE(U)=P(DT)xE(Ud)+P(UC)xE(Uu)
=0.2x12,000+0.8x16,000
=2.400+12.800
=15.200
So John Wants To Know The Total Value of The Car In Respect To The Possibility of A Broken Transmission
cars not damaged : $ 16,000 ; cars damaged : $ 12,000
ReplyDelete20% = 20/100 = 0,2
cars damaged 100% - 20% = 80% = 80/100 = 0,8
E(v) = (0.8 x 16,000) + (0.2 x 12,000) = 15,2
so the expected value of the car to smith is $ 15,2
Answer: Expected Value = E($) = Pr(X1) + (1 - Pr)(X2),
ReplyDeletewhere Pr is the probability of no transmission damage and Xi is the book value of the car without and with transmission damage, respectively.
E($) = .80(16,000) + .20(12,000)
= 12,800 + 2,400
= $15,200