In a valid probability distribution, each probability must be between 0 and 1,inclusive, and the probabilities must add up to 1. If a probability distribution5has probabilities1,112, and x, what is the value of X?

Given:
The probability distribution is given by,
[tex]\frac{5}{12},\frac{1}{6},\frac{1}{3},x[/tex]To find:
The value of x.
Explanation:
We know that,
The probabilities must add up to 1.
So, we write
[tex]\frac{5}{12}+\frac{1}{6}+\frac{1}{3}+x=1[/tex]Solving for x we get
[tex]\begin{gathered} x=1-\frac{5}{12}-\frac{1}{6}-\frac{1}{3} \\ x=\frac{12-5-2-4}{12} \\ x=\frac{1}{12} \end{gathered}[/tex]Therefore, the value of x is,
[tex]\frac{1}{12}[/tex]Final answer: Option D
The value of x is,
[tex]\frac{1}{12}[/tex]