## anonymous one year ago 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?

1. amistre64

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

2. anonymous

ummm only marriage? well how would know?

3. 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?

4. amistre64

3 drunks to 1 sober right?

5. anonymous

1/3

6. anonymous

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

7. anonymous

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

8. 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?

9. anonymous

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

10. 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.

11. 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.

12. anonymous

3/4 would be right

13. amistre64

yes

14. 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.

15. 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 $P(D|M)=\frac{n(D\cap M)}{n(M)}$

16. 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.

17. 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')

18. anonymous

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

19. amistre64

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

20. amistre64

no, D to D' is 3 to 1

21. amistre64

given M

22. 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.

23. 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.

24. 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.

25. 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

26. amistre64

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

27. amistre64

good luck :)

28. anonymous

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

29. amistre64

oh im interested :)

30. 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. :-)