Ok, so:
For Part A, we have: P(Z|A)=P(Z and A)/P(A)
And if we replace, we got:
P(Z|A) = (0.15)/(0.25) and this is equal to 0.6.
For Part B, we have: P(A|Z)=P(Z and A)/P(Z)
P(A|Z) = (0.15)/(0.73) and this is equal to 0.205.
The mean of the discrete distribution that models this situation is of 2.98.
<h3>What is the mean of a discrete distribution?</h3>
The expected value of a discrete distribution is given by the <u>sum of each outcome multiplied by it's respective probability</u>.
Considering the table, the probability distribution is given by:
Hence the mean is given by:
E(X) = 4 x 0.4 + 3 x 0.32 + 2 x 0.17 + 1 x 0.08 + 0 x 0.03 = 2.98.
More can be learned about the mean of a discrete distribution at brainly.com/question/27899440
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You would have to make the numbers have the same denominator. So you would have -15/20 + 4/20
Then you would have to put the sign of the higher number on it, So you would have -15/20 + 4/20 = -11/20
