Question

The probability distribution table below has some missing values. Find P(x=0)+P(x=3)0.90.050.10.5

Answer 1

The sum of the missing values is;

`P(X=0)+P(X=3)=0.1`

Step-by-step explanation:

Given the probability distribution table in the attached image.

`\begin{gathered} P(X=1)=0.6 \\ P(X=2)=0.3 \end{gathered}`

So, we have;

`\begin{gathered} P(X=0)+P(X=3)=1-(P(X=1)+P(X=2)) \\ P(X=0)+P(X=3)=1-(0.6+0.3) \\ P(X=0)+P(X=3)=1-(0.9) \\ P(X=0)+P(X=3)=0.1 \end{gathered}`

Therefore, the sum of the missing values is;

`P(X=0)+P(X=3)=0.1`