Concepcion Hale

2022-01-19

How do you determine the required value of the missing probability to make the following distribution a discrete probability distribution?
X P(x)
0 0.30
1 0.15
2 ?
3 0.20
4 0.15
5 0.05

braodagxj

Expert

P(2)=0.15
Explanation:
In any Prob. Dist., we must have, (1) $\sum P\left(x\right)=1,$ &, (2) each $P\left(x\right)\ge 0$.
Now, $\sum P\left(x\right)=1⇒P\left(0\right)+P\left(1\right)+P\left(2\right)+\dots +P\left(5\right)=1$
$⇒0.30+0.15+P\left(2\right)+0.20+0.15+0.05=1$
$⇒P\left(2\right)=0.15.$
We readily have, each $P\left(x\right)\ge 0.$

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