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An number cube (a fair die) is rolled 3 times. For each roll, we are interested in whether the roll comes up even or odd. An outcome is represented by a string of the sort "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 odd rolls in each outcome. For example, if the outcome is "ooo," then [tex]$N (ooo) = 3$[/tex]. Suppose that the random variable [tex][tex]$X$[/tex][/tex] is defined in terms of [tex]$N$[/tex] as follows: [tex]$X = 2N - N^2 - 2$[/tex]. The values of [tex][tex]$X$[/tex][/tex] are given in the table below.

\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
Outcome & oeo & eee & ooo & eoo & oee & eeo & ooe & eoe \\
\hline
Value of [tex]$X$[/tex] & -2 & -2 & -5 & -2 & -1 & -1 & -2 & -1 \\
\hline
\end{tabular}

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

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



Answer :

GinnyAnswer
Let's determine the probability distribution of the random variable [tex]\(X\)[/tex] by calculating the probabilities [tex]\(P(X=x)\)[/tex] for each unique value of [tex]\(X\)[/tex]. Given the outcome strings and their corresponding values of [tex]\(X\)[/tex], we need to find the unique values of [tex]\(X\)[/tex] and their associated probabilities.

First, identify the unique values, which are already given. These are: [tex]\(-5\)[/tex], [tex]\(-2\)[/tex], and [tex]\(-1\)[/tex].

Second, count the number of times each unique value appears to find the frequency. This will help us calculate the probabilities.

Here are the given outcomes and their corresponding values of [tex]\(X\)[/tex]:

[tex]\[ \begin{array}{|c|c|} \hline \text{Outcome} & \text{Value of } X \\ \hline \text{oeo} & -2 \\ \text{eee} & -2 \\ \text{ooo} & -5 \\ \text{eoo} & -2 \\ \text{oee} & -1 \\ \text{eeo} & -1 \\ \text{ooe} & -2 \\ \text{eoe} & -1 \\ \hline \end{array} \][/tex]

Now, count the occurrences of each value:
- [tex]\(X = -5\)[/tex]: 1 occurrence
- [tex]\(X = -2\)[/tex]: 4 occurrences
- [tex]\(X = -1\)[/tex]: 3 occurrences

There are a total of [tex]\(8\)[/tex] outcomes. Therefore, the probabilities for each unique value are calculated by dividing the frequency of each value by the total number of outcomes.

Calculate the probabilities:

[tex]\[ P(X = -5) = \frac{1}{8} = 0.125 \][/tex]

[tex]\[ P(X = -2) = \frac{4}{8} = 0.5 \][/tex]

[tex]\[ P(X = -1) = \frac{3}{8} = 0.375 \][/tex]

Now, we can fill in the probability distribution table for [tex]\(X\)[/tex]:

[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Value } x \text{ of } X & -5 & -2 & -1 \\ \hline P ( X = x ) & 0.125 & 0.5 & 0.375 \\ \hline \end{array} \][/tex]

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