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?

- anonymous

- chestercat

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- amistre64

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

- anonymous

ummm only marriage? well how would know?

- 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?

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## More answers

- amistre64

3 drunks to 1 sober right?

- anonymous

1/3

- anonymous

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

- anonymous

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

- 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?

- anonymous

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

- 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.

- 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.

- anonymous

3/4 would be right

- amistre64

yes

- 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.

- 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
\[P(D|M)=\frac{n(D\cap M)}{n(M)}\]

- 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.

- 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')

- anonymous

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

- amistre64

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

- amistre64

no, D to D' is 3 to 1

- amistre64

given M

- amistre64

|dw:1444340929233:dw|
given M, the probability of D is the number of D within M, divided by number of M
\[P(D|M)=\frac{n(D\cap M)}{n(M)}\]
as far as i can recall this is the basic definition of how to define probability. Number of items desired, out of number of items total.

- anonymous

I do understand that part. But I just don't see how they are getting anywhere close to the answers that they have given.

- amistre64

yeah, ive read over the question to see if there is some other way to interpret it ... i cant see the 3 to 1 as meaning anything other than the subset of marriage issues.
i cant see how to get one of the options either.

- anonymous

i don't see it either, i emailed the professor to see what he says, I just don't see what we are missing

- amistre64

if it was talking about the subset of drunks; it would read:
3 out of 4 marriages have drinking issues.

- amistre64

good luck :)

- anonymous

I was able to solve the problem if you are interested.

- amistre64

oh im interested :)

- anonymous

It was actually quite simple once I was able to get past my block of the proportion that was throwing me off. But anyways so the solution...
Like we know from the given: P[D] = 0.10, and the ratio P[M|D] = 3P[M|D'] (now to me isn't wasn't obvious at first bc I was thinking intersection of the two instead of conditional). And then of course what we want to find P[D|M]
Well Bayes Theorem states:
\[P(D|M)=\frac{ P(M|D)*P(D) }{ P(M|D)*P(D)+P(M|D')*P(D') }\]
Using the given information and substituting for P(M|D)
\[P(D|M)=\frac{ 3P(M|D')*P(D) }{ 3P(M|D')*P(D)+P(M|D')*P(D') }\]
And now we just factor out the P(M|D') from the numerator and denominator and throw in the given numbers
\[P(D|M)=\frac{ P(M|D') }{ P(M|D') } * \frac{ 3(.10) }{ (3(.10) + 0.9 } = \frac{ .30 }{ 1.2 }=0.25\]
And we are done. :-)

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