ValuexofX PXx 6 013 3 011 2 025 0 3 Fill in the values

ValuexofX

P=Xx

?6   =   0.13

?3  =    0.11

?2   = 0.25

0 = ?

3 =?

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

Solution

0.13+0.11+0.25 = 0.49

The sum of all probabilities must be 1, so we can distribute teh reaming 0.51 in any way we like, for example, we can have P(X=0) = 0.25, and P(X=3) = 0.26

Note that probability cannot be negative, and only lies between 0 and 1 (inclusive)