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January 11th, 2009, 10:20 PM
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A bag contains ten $1 bills, five $2 bills, four $5 bills, one $10 bill, one $20 bill, and one $100 bill. In order to play the game, which consists of drawing on bill randomly out of the bag, you must pay $20.
Q: A construct a probability distribution for this situation.
A: ????
Q: Find the expected value
A: I think, (1x10)+(2x5)+(5x4)+(10x1)+(20x1)+(100x1) = $7.27
Q:Is this game fair? if it is not fair, determine how much someone should have to pay in order for the game to be fair.
A: Well it should cost $7.27 for the game to be considered fair.
I dont know what to do on the first one, nor if i have done the 2nd one right, which therefore would make me do the last one wrong.
A: |
January 11th, 2009, 10:59 PM
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Quote:
Originally Posted by Bradley55  A bag contains ten $1 bills, five $2 bills, four $5 bills, one $10 bill, one $20 bill, and one $100 bill. In order to play the game, which consists of drawing on bill randomly out of the bag, you must pay $20.
Q: A construct a probability distribution for this situation.
A: ????
Q: Find the expected value Mr F says: The expected value of the amount of money pulled from the bag, I assume ....?
A: I think, (1x10)+(2x5)+(5x4)+(10x1)+(20x1)+(100x1) = $7.27 Mr F says: How did you get $7.27!? The left hand side adds up to 170 ....!
The answer (use the probability distribution - see how to get it below) is found using the definition: 170/22 = $7.73.
Q:Is this game fair? if it is not fair, determine how much someone should have to pay in order for the game to be fair.
A: Well it should cost $7.27 for the game to be considered fair.
I dont know what to do on the first one, nor if i have done the 2nd one right, which therefore would make me do the last one wrong.
A: | There are 22 bills. Ten of them are $1 bills. So Pr(X = $1) = 10/22. Calculate Pr(X = $2), Pr(X = $5) etc. in a similar way. The exhaustive set of probabilities is the probability distribution for the amount of money pulled from the hat.
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)
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January 11th, 2009, 10:59 PM
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(10/22)*(-19)+(5/22)*(-18)+(4/22)*(-15)+(1/22)*(-10)+(1/22)*(0)+(1/22)*(80) = $-12.27
so -12.27 is the expected value?
if so how do i do problem A?
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January 11th, 2009, 11:06 PM
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Quote:
Originally Posted by Bradley55 (10/22)*(-19)+(5/22)*(-18)+(4/22)*(-15)+(1/22)*(-10)+(1/22)*(0)+(1/22)*(80) = $-12.27
so -12.27 is the expected value?
if so how do i do problem A? | Expected value of what??
The expected value of money pulled from the bag is $7.73.
The expected value of money you win is 7.27 - 20 = - $12.27.
I have already said how to get the probability distribution. What part of what I said don't you understand?
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)
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January 11th, 2009, 11:41 PM
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so we know the expected value is -12.27
i got 3.86 for how much the game should cost for the game to be fair.
right?
i did (10/22)(x-1)+(5/22)(x-2)... so on and plugged that equation into y1. and x into y2, the intersected at 3.86,3.86 |
January 11th, 2009, 11:57 PM
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Quote:
Originally Posted by Bradley55 so we know the expected value is -12.27
i got 3.86 for how much the game should cost for the game to be fair.
right?
i did (10/22)(x-1)+(5/22)(x-2)... so on and plugged that equation into y1. and x into y2, the intersected at 3.86,3.86 | No.
For the game to be fair you require the expected value of the money you win to be zero.
Let the cost per game be C. Then you require 0 = 7.73 - C.
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)
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January 11th, 2009, 11:59 PM
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so expected value is -12.27 (all it said what to
and therefore the cost of the game should be $7.73 for it to be fair!
Woot
*edit*
b.) $-12.27
c.)$7.73
Corecto-mundo?
Last edited by Bradley55; January 12th, 2009 at 12:54 AM.
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