Fill in the [tex][tex]$P(X=x)$[/tex][/tex] values to give a legitimate probability distribution for the discrete random variable [tex][tex]$X$[/tex][/tex], whose possible values are [tex]1, 2, 4, 5[/tex], and [tex]6[/tex].

[tex]\[
\begin{tabular}{|c|c|}
\hline
\text{Value } x \text{ of } X & P(X=x) \\
\hline
1 & 0.21 \\
\hline
2 & 0.13 \\
\hline
4 & 0.11 \\
\hline
5 & \square \\
\hline
6 & \square \\
\hline
\end{tabular}
\][/tex]



Answer :

To complete the probability distribution for the discrete random variable [tex]\(X\)[/tex], we need to determine the probabilities [tex]\(P(X=5)\)[/tex] and [tex]\(P(X=6)\)[/tex].

Given the existing probabilities:
[tex]\[ P(X=1) = 0.21 \][/tex]
[tex]\[ P(X=2) = 0.13 \][/tex]
[tex]\[ P(X=4) = 0.11 \][/tex]

The sum of all probabilities must equal 1. Therefore, we set up the equation:
[tex]\[ P(X=1) + P(X=2) + P(X=4) + P(X=5) + P(X=6) = 1 \][/tex]
Substituting the known values:
[tex]\[ 0.21 + 0.13 + 0.11 + P(X=5) + P(X=6) = 1 \][/tex]

Now, let's calculate the sum of the known probabilities:
[tex]\[ 0.21 + 0.13 + 0.11 = 0.45 \][/tex]

Next, we find the remaining probability:
[tex]\[ 1 - 0.45 = 0.55 \][/tex]

Assuming that the probabilities [tex]\(P(X=5)\)[/tex] and [tex]\(P(X=6)\)[/tex] are equal (since no other information is given), we divide the remaining probability equally between them:
[tex]\[ P(X=5) = P(X=6) = \frac{0.55}{2} = 0.275 \][/tex]

Therefore, the values for the probability distribution are:
[tex]\[ P(X=5) = 0.275 \][/tex]
[tex]\[ P(X=6) = 0.275 \][/tex]

The completed probability distribution is:

[tex]\[ \begin{tabular}{|c|c|} \hline Value $x$ of $X$ & $P(X=x)$ \\ \hline 1 & 0.21 \\ \hline 2 & 0.13 \\ \hline 4 & 0.11 \\ \hline 5 & 0.275 \\ \hline 6 & 0.275 \\ \hline \end{tabular} \][/tex]

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