SOLUTION: Here are some probabilities estimated by a medical care study. The entries in the table are the probabilities that both of two events occur; for example, .321 is the probability th

Question 86761This question is from textbook Basic Practice of Statistics
: Here are some probabilities estimated by a medical care study. The entries in the table are the probabilities that both of two events occur; for example, .321 is the probability that a patient is under 65 years of age and the tests were done. The 4 probabilities in the table have a sum 1 beacuse the table lists all possible outcomes.
Tests Done?
Yes No
Age<65 0.321 0.124
Age>or=65 0.365 0.190
1) What is the probability that a pt. is under 65? That a pt. is 65 or over?
2) What is the probability that the tests were done for a pt.? That they were not done?
3) Are the events A= (the pt. was 65 or older) and B= (the tests were done independent? Were the tests omitted on older pts. more or less frequently than would be the case if testing were independent of age? This question is from textbook Basic Practice of Statistics

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Tests Done?
.........Yes... No......|Total
Age<65 0.321...0.124....|0.445
Age>=65 0.365...0.190....|0.555
-------- ----
Total...0.686...0.314.... 1.000
--------------------------
1) What is the probability that a pt. is under 65? Ans: 0.455
That a pt. is 65 or over ? Ans: 0.555
-----------------------------------
2) What is the probability that the tests were done for a pt.? Ans: 0.686
That they were not done? Ans: 0.314
---------------------------------
3) Are the events A= (the pt. was 65 or older) and B= (the tests were done independent?
P(A)*P(B) = 0.555*0.686 = 0.38073
P(A|B) = 0.365
Ans: Since these are not the same the A and B are dependent.
--------------------
Were the tests omitted on older pts. more or less frequently than would be the case if testing were independent of age?
Ans: Yes, the percentage would be up to 38% instead of 36.5%
=============
Cheers,
Stan H.

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