Posted by **B.B.** on Monday, August 10, 2009 at 10:17am.

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.

- Math -
**bobpursley**, Monday, August 10, 2009 at 10:30am
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

- Math -
**B.B.**, Monday, August 10, 2009 at 10:39am
So then, my answer should be this:

13,333.33333+1+16.66666667= 13,351 is this correct?

Thanks.

- Math -
**Marth**, Monday, August 10, 2009 at 11:39am
"Expected Value is in dollars."

Do not forget your units. 13,351 is the correct value, but your answer should be $13,351.00

- Math -
**B.B.**, Monday, August 10, 2009 at 11:46am
Thanks.

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