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?
Answer: 1/3 but I don't think that this correct.
Could someone please help me with this problem?
Thanks.
OF course it is not correct. Expected Value is in dollars.
EV=1/3(40,000) + 1/3 (3) + 1/3 (50)
http://en.wikipedia.org/wiki/Expected_value
So then, my answer should be this:
13,333.33333+1+16.66666667= 13,351 is this correct?
Thanks.
"Expected Value is in dollars."
Do not forget your units. 13,351 is the correct value, but your answer should be $13,351.00
Thanks.
To calculate the expected value, we need to multiply the probability of each outcome by its corresponding value, and then sum them up.
In this case, the probability of selecting any particular door is 1/3, since all doors have an equal probability of being selected. Let's denote the doors by A, B, and C, with their respective prizes as follows:
A: $40,000 foreign car
B: $3 silly straw
C: $50 mathematics textbook
Now, let's calculate the expected value step by step:
Expected value = (probability of A) * (value of A) + (probability of B) * (value of B) + (probability of C) * (value of C)
Expected value = (1/3) * ($40,000) + (1/3) * ($3) + (1/3) * ($50)
Expected value = ($40,000/3) + ($3/3) + ($50/3)
Expected value = $13,333.333 + $1 + $16.667
Expected value = $13,350
Therefore, the expected value of the selection is $13,350.
Regarding your question about the answer being 1/3, it seems incorrect. The answer should be the calculated expected value, which in this case is $13,350.