Question

random variables and distributions
discrete probability distribution: basic
fill in the $p(x = x)$ values to give a legitimate probability distribution for the discrete random variable $x$, whose possible values are 1, 3, 4, 5, and 6.
value $x$ of $x$ $p(x = x)$
1 0.25
3
4 0.30
5
6 0.12

Explanation:

Step1: Recall the property of probability distribution

The sum of all probabilities in a discrete probability distribution is \(1\). Let \(P(X = 3)=a\) and \(P(X = 5)=b\). Then \(0.25+a + 0.30+b+0.12=1\).

Step2: Simplify the equation

Combine like - terms: \(a + b=1-(0.25 + 0.30+0.12)\). Calculate \(1-(0.25 + 0.30+0.12)=1 - 0.67=0.33\).

Step3: Assume a common case (uniform for \(x = 3\) and \(x = 5\) if no other constraints)

If we assume \(a=b\) (since no other information is given to distinguish between the probabilities of \(x = 3\) and \(x = 5\)), then \(a=b=\frac{0.33}{2}=0.165\).

Answer:

\(P(X = 3)=0.165\) and \(P(X = 5)=0.165\)