Example Question with answer, My problem is at the bottom.

Fill in theP(X = x)values in the table below to give a legitimateprobability distributionfor the discreterandom variableX, whose possible values are6,2,3,5,and6.

Value x of X: -6, 2, 3, 5, 6

P(X=x): 0.25, 0.21, 0.13, __, __

EXPLANATION

The values that the discrete random variableX

may take are6,2,3,5,and6.Thus, it must be that P(X= -6)+P(X=2)+P(X=3)+PX=5)+P(X=6) = 1, with 0 < equal to sign P(X=x) < equal to sign 1 for x = -6, 2, 3, 5, 6

From the information given, we have 0.25+0.21+0.13+P(X=5)+P(X=6) = 1,

which means that P(X=5)+P(X=6) = 1-(0.25+0.21+0.13) =0.41.

To complete the probability distribution, then, we must chooseP (X=5) and P(X=6)to be any numbers from0to1such that P(X-5)+P(X=6) = 0.41

.

For example, we could chooseP(X=5) = 0.17 and P(X=6) = 0.24

One correct answer is:

ANSWER

Value x of X: -6, 2, 3, 5, 6

P (X=x): 0.25, 0.21, 0.13, 0.17, 0.24

My problem that I am having trouble with:

Fill in theP (X=x)values in the table below to give a legitimateprobability distributionfor the discreterandom variableX, whose possible values are0,2,4,5,and6.

Value x of X: 0, 2, 4, 5, 6

P(X=x): 0.30, __, 0.12, __, 0.28