Part (a)
Here is what the spinner might look like
Regions A and D are red
Regions C and F are yellow
Regions E and G are purple
Region B is blue
The order and placement of each region doesn't matter.
We have these four events:
landing on a red region
landing on a blue region
landing on a yellow region
landing on a purple region
Red, yellow, and purple each have probability of 2/7 because there are 2 of each color out of 7 total.
Blue has probability 1/7.
landing on blue gives 3 points.
landing on yellow gives 1 point.
landing on purple gives 0 points.
landing on red gives -1 point (i.e. 1 point is taken away).
Table
| Color | X | P(X) |
| Red | -1 | 2/7 |
| Purple | 0 | 2/7 |
| Yellow | 1 | 2/7 |
| Blue | 3 | 1/7 |
X = points awarded after each spin
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Part (b)
Let's extend the table we made in part (a).
We'll attach another column labeled X*P(X).
As the new column label indicates, we'll multiply each X with their corresponding P(X) value.
Example:
X = 3 (blue)
P(X) = 1/7
X*P(X) = 3*(1/7) = 3/7
| Color | X | P(X) | X*P(X) |
| Red | -1 | 2/7 | -2/7 |
| Purple | 0 | 2/7 | 0 |
| Yellow | 1 | 2/7 | 2/7 |
| Blue | 3 | 1/7 | 3/7 |
Then add up the values in that X*P(X) column to find the expected value.
-2/7 + 0 + 2/7 + 3/7
-2/7 + 0/7 + 2/7 + 3/7
(-2+0+2+3)/7
3/7
Optionally you could use a calculator to find that 3/7 = 0.428571 approximately.
Answer as a fraction = 3/7
Answer as a decimal = 0.428571 (approximate)
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Part (c)
The nonzero expected value (from part b) means the game is not mathematically fair.
A fair game is when the expected value is zero. Neither side wins nor loses.
Let's go back to this expression
(-2+0+2+3)/7
which was the third step in the scratch work computing the expected value (see part b)
The 3 at the end of (-2+0+2+3)/7 represents the points you get on that blue space.
Let's say we replace 3 with x to get (-2+0+2+x)/7
Set that equal to zero and solve for x.
(-2+0+2+x)/7 = 0
(0+x)/7 = 0
x/7 = 0
x = 7*0
x = 0
Therefore, one adjustment you can make is to have the blue space award 0 points (instead of 3 points).
Doing this will make the game fair.
Other adjustments could be made.
I'll let the student create the new table for this scenario, and recalculate the expected value based on that new table.
You should get an expected value of zero.