Question 1197919: James designed a game that uses a spinner with twelve equal sectors numbered 1 to 12. If the spinner stops on an odd number, a player gains that number of points. However, if the spinner stops on an even number, the player losses that number of points.
a) Show the probability distribution of points in this game (table only)
b) What is the expected number of points per spin?
c) Is the game fair? Explain
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
| Spinning a... | Points Won | Probability | | 1 | 1 | 1/12 | | 2 | -2 | 1/12 | | 3 | 3 | 1/12 | | 4 | -4 | 1/12 | | 5 | 5 | 1/12 | | 6 | -6 | 1/12 | | 7 | 7 | 1/12 | | 8 | -8 | 1/12 | | 9 | 9 | 1/12 | | 10 | -10 | 1/12 | | 11 | 11 | 1/12 | | 12 | -12 | 1/12 |
X = number of points won
P(X) = probability of winning X points
| X | P(X) | | 1 | 1/12 | | -2 | 1/12 | | 3 | 1/12 | | -4 | 1/12 | | 5 | 1/12 | | -6 | 1/12 | | 7 | 1/12 | | -8 | 1/12 | | 9 | 1/12 | | -10 | 1/12 | | 11 | 1/12 | | -12 | 1/12 |
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Part (b)
Form a third column labeled X*P(X)
As you can probably guess, this column represents us multiplying each X and P(X) value
Eg: -2*(1/12) = -2/12
I'll leave the fractions un-reduced
| X | P(X) | X*P(X) | | 1 | 1/12 | 1/12 | | -2 | 1/12 | -2/12 | | 3 | 1/12 | 3/12 | | -4 | 1/12 | -4/12 | | 5 | 1/12 | 5/12 | | -6 | 1/12 | -6/12 | | 7 | 1/12 | 7/12 | | -8 | 1/12 | -8/12 | | 9 | 1/12 | 9/12 | | -10 | 1/12 | -10/12 | | 11 | 1/12 | 11/12 | | -12 | 1/12 | -12/12 |
Then we add up everything in the X*P(X) column.
We'll add the numerators
1+(-2)+3+(-4)+5+(-6)+7+(-8)+9+(-10)+11+(-12) = -6 which is the numerator in the answer before reducing.
Adding up all the fractions in the X*P(X) column leads to -6/12 = -1/2
The player expects to lose, on average, 1/2 = 0.5 points per game.
Answer: -1/2 = -0.5
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Part (c)
The game is not fair because the expected value (in part b) is not zero.
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