Question

Fill in the values to give a legitimate probability distribution for the discrete random variable , whose possible values are , , , , and .

Answer 1

We must complete the table of probabilities such that the probability distribution is legitimate.

Because all the possible values of the variables are 0, 3, 4, 5 and 6, the sum of the probability of each number must sum up 1:

`P(x=0)+P(x=3)+P(x=4)+P(x=5)+P(x=6)=1.`

Replacing the values of the table, we have:

`\begin{gathered} 0.12+P(x=3)+P(x=4)+0.24+0.20=1, \\ P(x=3)+P(x=4)=1-0.12-0.24-0.20, \\ P(x=3)+P(x=4)=0.44. \end{gathered}`

From the last equation, we see that to have a legitimate probability, we must select values of P(x = 3) and P(x = 4) such that they sum up 1. One possible pair of values is:

• P(x = 3) = 0.2,

,

• P(x = 4) = 0.24.

Answer

One possible pair of values is:

• P(x = 3) = 0.2,

,

• P(x = 4) = 0.24.