Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

Probability Help!!! The probability that a randomly chosen male has a drinking problem is 0.10 . Males who have drinking problems are three times as likely to have marriage problems as those who do not have a drinking problem. What is the conditional probability that a male has a drinking problem, given that he has marriage problems?

OpenStudy (amistre64):

spose we put all the people with marriage problems into one room ... how many people would be in the room?

OpenStudy (anonymous):

ummm only marriage? well how would know?

OpenStudy (amistre64):

the actual number of people in the room is not important .. it can be any number we chose. but given, that we have a room full of marriage issues ... what is the ratio of drunks to sobers?

OpenStudy (amistre64):

3 drunks to 1 sober right?

OpenStudy (anonymous):

1/3

OpenStudy (anonymous):

So I figured it was P(M and D) = 3 P(M and D')

OpenStudy (anonymous):

and P(D) = 0.10 and we want to find P(D | M)

OpenStudy (amistre64):

well, 3 to 1 sounds better to me ... drunks to sober so lets say there are 4 people in the room. 3 of them are drunks. whats that give us as a probability?

OpenStudy (anonymous):

you asking whats the probability that out of those 4 they have marriage issues or just out of the 3

OpenStudy (amistre64):

P(D) is not relevant to the question. i believe it is just added info to see if you know what you are looking for.

OpenStudy (amistre64):

spose we have 4 people in the room ... given this set of people, what is the probability that we can pick a drunk? in other words: what is the probability of picking a drink; given a room full of people with marriage issues.

OpenStudy (anonymous):

3/4 would be right

OpenStudy (amistre64):

yes

OpenStudy (anonymous):

given that the answer choices are .05, .10, .15, .20, .25.... That couldn't be right... i see what you are saying as far as this logic goes... but it feels like something is being left out.

OpenStudy (amistre64):

hmm, lets see if im reading it correctly, or at least as correct as what makes sense :) The probability that a randomly chosen male has a drinking problem is 0.10 . Males who have drinking problems are three times as likely to have marriage problems as those who do not have a drinking problem. **** this is a 3 drunks to 1 sober ratio right? ****So of the subset of 'has marriage problems', we would expect that 3 out of 4 people in the room are drunks. What is the conditional probability that a male has a drinking problem, given that he has marriage problems? \[P(D|M)=\frac{P(D\cap M)}{P(M)}\] or written another way \[P(D|M)=\frac{n(D\cap M)}{n(M)}\]

OpenStudy (amistre64):

we arent given the number (or probability) of marriage issues. It is just hinted that the ratio is 3 to 1, or 3n to n right? i cant see the error in my thoughts.

OpenStudy (anonymous):

Well to me, and what I can't decide which I have a big feeling is part of the problem, well cause every problem in this text i have ever come across uses absolutely every piece of information... is the 3 to 1 part... now I can't figure out if that means P(M and D) = 3 P(M and D') or P(M | D) = 3 P(M | D')

OpenStudy (anonymous):

And you use that info along with the P(D) = .10 and P(D') = .9 to solve for P(D | M)

OpenStudy (amistre64):

|dw:1444340496158:dw| maybe DnM = .75?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!