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An outcome is represented by a string such as "oee" (meaning an odd number on the first roll, an even number on the second roll, and an even number on the third roll).

For each outcome, let [tex]$N$[/tex] be the random variable counting the number of even rolls in each outcome. For example, if the outcome is "eee", then [tex]$N(\text{eee}) = 3$[/tex]. Suppose that the random variable [tex][tex]$X$[/tex][/tex] is defined in terms of [tex]$N$[/tex] as follows: [tex]$X = 2N - N^2 - 3$[/tex]. The values of [tex][tex]$X$[/tex][/tex] are given in the table below.

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
Outcome & eeo & eoo & eee & ooe & ooo & eoe & oeo & oee \\
\hline
Value of $X$ & -3 & -2 & -6 & -2 & -3 & -3 & -2 & -3 \\
\hline
\end{tabular}
\][/tex]

Calculate the probabilities [tex]$P(X=x)$[/tex] of the probability distribution of [tex]$X$[/tex]. First, fill in the first row with the values of [tex][tex]$X$[/tex][/tex]. Then fill in the appropriate probabilities in the second row.

[tex]\[
\begin{tabular}{|l|c|c|c|}
\hline
Value $x$ of $X$ & $\square$ & $\square$ & $\square$ \\
\hline
$P(X = x)$ & $\square$ & $\square$ & $\square$ \\
\hline
\end{tabular}
\][/tex]

Sagot :

GinnyAnsweridn
Let's break down the solution to this problem step-by-step.

### Step 1: Identify the outcomes and their corresponding values of [tex]\( X \)[/tex]

The outcomes and the corresponding values of the random variable [tex]\( X \)[/tex] are given:
- eeo: [tex]\( X = -3 \)[/tex]
- eoo: [tex]\( X = -2 \)[/tex]
- eee: [tex]\( X = -6 \)[/tex]
- ooe: [tex]\( X = -2 \)[/tex]
- ooo: [tex]\( X = -3 \)[/tex]
- eoe: [tex]\( X = -3 \)[/tex]
- oeo: [tex]\( X = -2 \)[/tex]
- oee: [tex]\( X = -3 \)[/tex]

### Step 2: List the given values of [tex]\( X \)[/tex]

The given values of [tex]\( X \)[/tex] for each outcome are:
[tex]\[ [-3, -2, -6, -2, -3, -3, -2, -3] \][/tex]

### Step 3: Calculate the frequency of each value of [tex]\( X \)[/tex]

Next, we determine the frequency (how many times each unique value appears in the list):

- [tex]\( X = -6 \)[/tex] appears 1 time
- [tex]\( X = -3 \)[/tex] appears 4 times
- [tex]\( X = -2 \)[/tex] appears 3 times

### Step 4: Calculate the total number of outcomes

There are a total of 8 outcomes.

### Step 5: Calculate the probability distribution [tex]\( P(X=x) \)[/tex]

The probability of each value of [tex]\( X \)[/tex] is given by the frequency of that value divided by the total number of outcomes:

- [tex]\( P(X = -6) \)[/tex] : [tex]\(\frac{1}{8} = 0.125\)[/tex]
- [tex]\( P(X = -3) \)[/tex] : [tex]\(\frac{4}{8} = 0.5\)[/tex]
- [tex]\( P(X = -2) \)[/tex] : [tex]\(\frac{3}{8} = 0.375\)[/tex]

### Step 6: Organize the values and probabilities in a table

Now, filling in the given table format:

[tex]\[ \begin{tabular}{|l|c|c|c|} \hline Value \( x \) of \( X \) & -6 & -3 & -2 \\ \hline \( P(X = x) \) & 0.125 & 0.5 & 0.375 \\ \hline \end{tabular} \][/tex]

So, the completed table is:

[tex]\[ \begin{tabular}{|l|c|c|c|} \hline Value \( x \) of \( X \) & -6 & -3 & -2 \\ \hline \( P(X = x) \) & 0.125 & 0.5 & 0.375 \\ \hline \end{tabular} \][/tex]

This table shows the probability distribution for the random variable [tex]\( X \)[/tex].
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