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Answer to Question #6887 in Statistics and Probability for Misael

Question #6887
The probability that a married man watches a certain television show is 0.4 and the probability that a married woman watches the show is 0.5. The probability that a man watches the show, given that his wife does, is 0.7. Find the probability that
a) a married couple watches the show
b) a wife watches the show given that her husband does;
c) at least 1 person of a married couple will watch the show.
Expert's answer
a) a married couple watches the show

P(man|woman)=[P(man AND woman)]/[P(woman)]
Therefore (man AND woman) = P(man|woman)*P(woman) = 0.7*0.5 = 0.35.

b) a wife watches the show given that her husband does:

P(woman|man) = P(man AND woman)/P(man) = 0.35/0.4 = 0.8750

c) at least 1 person of a married couple will watch the show.



Need to figure out
P(both watch) = 0.35
P(man watches and woman does not) =P(man watches) - P(both watch)=0.40-0.35 = 0.05
P(woman watches and man does not) =P(woman watches) - P(both watch)= 0.50-0.35 = 0.15
P(at least 1 person of a couple will watch) =P(man watches and woman does not)+ P(woman watches and man does not)+P(both watch)=0.05+0.15+0.35 = 0.55

Therefore P(at least 1 person of a couple will watch)= 0.55.

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Comments

Assignment Expert
03.09.14, 20:56

Dear John.
 Thank you for correcting us. You have used another formula for probability of union of two events. Our solution applies formula for probability of union of two mutually exclusive events.

John
02.09.14, 00:36

Hi, I think c) is not right. 0.55 is the probability that at least one person watch the program (man, woman or both). Also you can write this like this: P(man or woman)= P(man)+P(woman)-P(both)= 0.4+0.5-0.35=0.55

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