On a TV game show, the contestant is asked to select a
door and then is rewarded with the prize behind the door
selected. If the doors can be selected with equal probability,
what is the expected value of the selection if the
three doors have behind them a $40,000 foreign car, a $3
silly straw, and a $50 mathematics textbook?
Each door has a prob of 1/3
expected value
= 1/3( 40000 + 3 + 50 ) = 13351
How did you get this answer I want to make sure im doing this correctly, thank you.
expected return on picking first door
= (1/3)(4000)
expected return on second door
= (1/3)(3)
expected return on third door
= (1/3)50)
so expected return on whole event
= (1/3)(4000) + (1/3)(3) + (1/3)(50)
= (1/3)( 40000 + 3 + 50 ) = 13351 as above
To calculate the expected value of the selection, you need to multiply each possible outcome by its corresponding probability and then sum them up. In this case, there are three doors with different prizes:
1. $40,000 foreign car
2. $3 silly straw
3. $50 mathematics textbook
Since the doors can be selected with equal probability, each door has a 1/3 chance of being chosen. Let's calculate the expected value step by step:
1. The probability of selecting the $40,000 car is 1/3. Multiplying the value of the car by its probability gives us: (1/3) * $40,000 = $13,333.33
2. The probability of selecting the $3 silly straw is also 1/3. Multiplying the value of the silly straw by its probability gives us: (1/3) * $3 = $1
3. The probability of selecting the $50 mathematics textbook is again 1/3. Multiplying the value of the textbook by its probability gives us: (1/3) * $50 = $16.67
Now, sum up these values: $13,333.33 + $1 + $16.67 = $13,350
Therefore, the expected value of the selection is $13,350.