Explain conditional probability distribution.
Presume that during a basketball practice. Joe tries three shots from the free throw line, presume further that he has a probability equal to 0.6 of making each shot as well as that the attempts are independent. Let X be the number of baskets he makes in the first two shots also let Y be the number of baskets he makes in the last two shots. The marginal density for X as well as Y are each binomial with n = 2 and p = 0.6. The subsequent table shows the assignment of the probabilities.
|
Outcome
|
HHH
|
HHM
|
HMH
|
HMM
|
MHH
|
MHM
|
MMH
|
MMM
|
|
Probability
|
(0.6)3
|
(0.6)2 (0.4)
|
(0.6)2(0.4)
|
(0.6) (0.4)2
|
(0.6)2(0.4)
|
(06) (0.4)2
|
(06) (0.4)2
|
(0.4)3
|
|
Value for X
|
2
|
2
|
1
|
1
|
1
|
1
|
0
|
0
|
|
Value for Y
|
2
|
1
|
1
|
0
|
2
|
1
|
1
|
0
|
Please note that the probability values for the events {2. 0} and {0. 2} aren't defined and hence each is assigned to be equal to zero for this problem.
1) Complete the table showing the joint distribution function for X as well as Y for this problem using information shown in the table above
2) Define the marginal densities of X and Y
3) Define the conditional probability f (X | Y= l) and hence compute P(X = 0 | Y = l)
4) Define the conditional probability f (X = 2 | Y) and hence compute P(X = 21 Y = 0)