To determine which of the offered distributions represents a valid probability distribution, we need to check each distribution against the criteria for a valid probability distribution:
1. All probabilities must be between 0 and 1 (inclusive).
2. The sum of all probabilities must equal 1.
Let's analyze each distribution:
Probability Distribution A:
[tex]\[
\begin{aligned}
&\begin{array}{|c|c|}
\hline
X & P(x) \\
\hline
1 & -0.14 \\
\hline
2 & 0.6 \\
\hline
3 & 0.25 \\
\hline
4 & 0.29 \\
\hline
\end{array}
\end{aligned}
\][/tex]
1. The probability at [tex]\(X=1\)[/tex] is [tex]\(-0.14\)[/tex], which is less than 0. Hence, one of the probabilities is not in the range [0, 1].
2. The sum of the probabilities is:
[tex]\[
-0.14 + 0.6 + 0.25 + 0.29 = 1.0
\][/tex]
Although the sum is 1, the presence of a negative probability makes this distribution invalid.
Probability Distribution B:
[tex]\[
\begin{aligned}
&\begin{array}{|c|c|}
\hline
X & P(x) \\
\hline
1 & 0 \\
\hline
2 & 0.45 \\
\hline
3 & 0.16 \\
\hline
4 & 0.39 \\
\hline
\end{array}
\end{aligned}
\][/tex]
1. All probabilities ([tex]\(0, 0.45, 0.16, 0.39\)[/tex]) are within the range [0, 1].
2. The sum of the probabilities is:
[tex]\[
0 + 0.45 + 0.16 + 0.39 = 1
\][/tex]
Both conditions are met, so this distribution is valid.
Probability Distribution C:
[tex]\[
\begin{aligned}
&\begin{array}{|c|c|}
\hline
X & P(x) \\
\hline
1 & 0.45 \\
\hline
2 & 1.23 \\
\hline
3 & -0.87 \\
\hline
\end{array}
\end{aligned}
\][/tex]
1. The probability at [tex]\(X=2\)[/tex] is [tex]\(1.23\)[/tex], which is greater than 1, and the probability at [tex]\(X=3\)[/tex] is [tex]\(-0.87\)[/tex], which is less than 0. Hence, there are probabilities not in the range [0, 1].
2. The sum of the probabilities is:
[tex]\[
0.45 + 1.23 - 0.87 = 0.81
\][/tex]
Given that there are values out of the [0, 1] range and the sum is not 1, this distribution is invalid.
Conclusion: Among the given distributions, only Probability Distribution B is valid.
