## INT 3 years ago The probability that a woman exposed to German measles will contract the disease is 0.2. if exposed to risk while pregnant, the probability is 0.1 that the child will suffer from a particular serious birth defect. Otherwise, the probability of such a defect is only 0.01. A child was born with this birth defect. What is the probability that the mother suffered from German measles while pregnant? The probability that a woman exposed to German measles will contract the disease is 0.2. if exposed to risk while pregnant, the probability is 0.1 that the child will suffer from a particular serious birth defect. Otherwise, the probability of such a defect is only 0.01. A child was born with this birth defect. What is the probability that the mother suffered from German measles while pregnant? @Mathematics

1. satellite73

baye's formula right?

2. INT

honestly, I have no idea I think it might be

3. satellite73

or at least we can use baye's formula, but sometimes the formula can be a bit confusing, because you forget what is conditioned on what. so we can do this the boneheaded way (which is how i would do it) and then maybe even baye's formula will make sense. let me write it on paper so i don't make a mistake.

4. satellite73

now i am confused because we don't know the probability that a woman is exposed to the measles. so i have to thing why that is not important.

5. AnwarA

The problem I'm facing is how to finding the probability that a child will suffer from a particular serious birth defect regardless weather his mother was exposed to German measles or not. Using Bayes' formula as satellite said $$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$ with A being the event that the woman was exposed to the German measles while pregnant and B the event that the child will suffer from a particular serious birth defect. We want to find $$P(A|B)$$. We have $$P(B|A)=0.1 \text{ and } P(A)=0.2$$. I think that $$P(B)=0.2\times0.1+0.8\times0.01=0.28$$. If that is right then $$P(A|B)=\frac{5}{7}$$.

6. satellite73

i am confused as to why if A is the even the woman was exposed while pregnant then $P(A)=0.2$

7. satellite73

*event

8. satellite73

this is what is bothering me, that we have no idea what $P(A)$ is

9. INT

according to the answer key, 5/7 is the correct answer. But I dont quite get how you did it.

10. satellite73

i can show you where the 5/7 comes from but i am not clear at all why it is right, so i will be quiet and let anwar explain

11. AnwarA

You made me confused too @satellite. :D

12. satellite73

i am having a problem that you don't know what the probability is a woman is exposed to the disease, only what the probability is that she gets is if she is exposed. so if you put A = is exposed and B = has the disease then you know $P(B|A)=0.2$ but you do not know $P(A)$

13. satellite73

if you put C = child has the defect, and you want $P(C)$ (sorry i changed your notation) then what you want to use is $P(C)=P(C|A)P(A)+P(C|A^c)P(A^c)$ but you don't know either of these. what you used is

14. jprahman

Well, $P(B) = P(A)0.2$ with B representing infection and A representing exposure. After all you can be infected if you aren't exposed. So now $5*P(B) = P(A)$

15. jprahman

Can't be infected.

16. AnwarA

I guess there could be something missing in the question. If my answer was right then the 0.2 has to be the probability if the woman was pregnant. Could you check again if the question says anything about this?

17. satellite73

i get what you are saying, and i realize you cannot get the disease if you are not exposed, but suppose say 100% of the women are exposed.

18. INT

The question typed is all that is given.

19. AnwarA

Hmm, yeah I agree with what you're saying @satellite.

20. INT

21. tcln1456

I see how it works: P(B|A)= P(A n B)/P(A) P(A|B)= P(A n B)/P(B) so P(A)P(B|A)= P(A n B) and you substitute that into P (A|B) = [P(A)P(B|A)]/P(B)

22. tcln1456

then it's plug and chug

23. satellite73

lets do an example with specific numbers and see if it works. if it does i will be silent. otherwise i think there is something wrong here. suppose there are 1,000,000 women total and all are exposed. then .2% of 1,000,000 is 2,000 so 2,000 women have the disease and 998,000 don't. now the child has the defect. it is certain that the woman was exposed (because they all were) what is the probability the mother had the disease? in this case the child having the defect is predictive of nothing, because all the mothers were exposed. so the probability the mother has the disease is exactly the same as the ratio of those that have it to the total.

24. satellite73

now let us assume no women were exposed. now the probability that the mother has the disease is 0

25. satellite73

i think some other assumption is being made here that is not stated, but i don't want to confuse the issue, so if 5/7 is the answer let it be.

26. satellite73

ok got it, although this is not the answer to the question as stated

27. satellite73

there is something wrong with the question, that if for sure. it is not "the probability that an exposed woman has measles is .2" but rather "the probability that a woman has the measles is .2" and that is a totally different thing

28. satellite73

so now we can do the problem with baye's formula, or even just with simple numbers. answar wrote out baye's i will write it with numbers (always a good way to proceed if the formula is confusing) lets imagine there are exactly 1000 pregnant women, and 20% = 200 have the measles and so 800 do not. of the 800 that do not, .01 * 800 = 8 will give birth to a child with the defect, and of the 200 who do, .1*200 = 20 will there are 28 children with the defect. 20 of those came from women with the disease 20/28 = 5/7 meaning that if the child has the disease the probability that the mother had the measles is 5/7 but this is an answer to what seems to me to be a very different question