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A game at the fair involves a wheel with seven sectors. Two of the sectors are red, two of the sectors are purple, two of the sectors are yellow, and one sector is blue.

Landing on the blue sector will give 3 points, landing on a yellow sector will give 1 point, landing on a purple sector will give 0 points, and landing on a red sector will give -1 point.

a. Let [tex]$X$[/tex] be the points you have after one spin. Fill out the missing values in the table.

\begin{tabular}{|c|c|c|c|c|}
\hline
[tex]$x_i$[/tex] & 3 & 1 & 0 & -1 \\
\hline
Frequency & 1 & 2 & 2 & 2 \\
\hline
[tex][tex]$P(x_i)$[/tex][/tex] & [tex]\(\frac{1}{7}\)[/tex] & [tex]\(\frac{2}{7}\)[/tex] & [tex]\(\frac{2}{7}\)[/tex] & [tex]\(\frac{2}{7}\)[/tex] \\
\hline
\end{tabular}

b. If you take one spin, what is your expected value?



Answer :

GinnyAnswer
Let's solve the problem step by step:

### a. Fill out the missing values in the table

We are given that there are four possible outcomes for the points you can get after one spin, along with their associated probabilities:

- Landing on the blue sector gives 3 points.
- Landing on a yellow sector gives 1 point.
- Landing on a purple sector gives 0 points.
- Landing on a red sector gives -1 point.

Given that there are seven sectors in total:

- One sector is blue, so the probability of landing on blue (3 points) is [tex]\( \frac{1}{7} \)[/tex].
- Two sectors are yellow, so the probability of landing on yellow (1 point) is [tex]\( \frac{2}{7} \)[/tex].
- Two sectors are purple, so the probability of landing on purple (0 points) is [tex]\( \frac{2}{7} \)[/tex].
- Two sectors are red, so the probability of landing on red (-1 point) is [tex]\( \frac{2}{7} \)[/tex].

We can place these values into a table:

[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline $x_{i}$ & 3 & 1 & 0 & -1 \\ \hline $P(x_{i})$ & \frac{1}{7} & \frac{2}{7} & \frac{2}{7} & \frac{2}{7} \\ \hline \end{tabular} \][/tex]

### b. Calculate the expected value

The expected value [tex]\( E(X) \)[/tex] is calculated using the formula:

[tex]\[ E(X) = \sum_{i} x_{i} \cdot P(x_{i}) \][/tex]

Substituting the values from the table into the formula:

[tex]\[ E(X) = 3 \cdot \frac{1}{7} + 1 \cdot \frac{2}{7} + 0 \cdot \frac{2}{7} + (-1) \cdot \frac{2}{7} \][/tex]

Calculating step-by-step:

[tex]\[ E(X) = \frac{3}{7} + \frac{2}{7} + 0 + \frac{-2}{7} \][/tex]
[tex]\[ E(X) = \frac{3}{7} + \frac{2}{7} - \frac{2}{7} \][/tex]
[tex]\[ E(X) = \frac{3}{7} \][/tex]

Thus, the expected value is:

[tex]\[ E(X) = 0.42857142857142855 \][/tex]

So, the step-by-step solution to the problem is:

### Solution:

a. The filled-out table is:
[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline $x_{i}$ & 3 & 1 & 0 & -1 \\ \hline $P(x_{i})$ & \frac{1}{7} & \frac{2}{7} & \frac{2}{7} & \frac{2}{7} \\ \hline \end{tabular} \][/tex]

b. The expected value if you take one spin is:
[tex]\[ 0.42857142857142855 \][/tex]

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