I'm having some trouble with these...
A test for Hepatitis A is 98% accurate: if a person has Hepatitis, the test will come out "positive" with 0.98 probability, and if a person doesn't have Hepatitis, the test will come out "negative" with 0.98 probability. The percentage of the population that has Hepatitis A is 0.005. Based on these probabilities, if a person tests positive for Hepatitis A, what is the probability that the person actually has Hepatitis A?
I got the answer for this by doing 0.98 * .005 = .0049, but it seems too easy of a calculation. Did I do this correctly?
The next problem says
Harry is an employee at a golf resort that has a couple of different courses. He's told by his owner to cut 10 greens and that he should spend only one hour on each green. For a par 3 green there is a 0.80 probability that he will cut it in one hour. For a par 5 green there is a 0.60 probability that he will have it cut in one hour. If his job is to prepare six par 3 greens and four par 5 greens, what is the probability that he will have at least 9 of them cut in the time given?
I tried a combinations of things for this problem like multiplying .80 with itself 6 times and adding that with (.60 * .60 * .60 * .60).It doesn't feel right though. Please get me headed in the direction
let H be that the person has Hepatitis and let T be that the person test positive of HepatitisOriginally Posted by DAI&DF
you want P(H|T)
=P(H and T)/P(T)
=P(T|H)*P(H)/P(T)
and
P(T)=P(T and H or T and H')
=P(T and H) + P(T and H')
=P(T|H)*P(H)+P(T|H')*P(H')
therefore
P(H|T)=P(T|H)*P(H)/[P(T|H)*P(H)+P(T|H')*P(H')]
If you look in our Examples section, you'll find a post on Bayes' Theorem and contingency tables, which is pretty much the same problem as your hepatitis example.
Thanks for replying Martingale and John M. I think I understand how to do it now. But is it ok if the problem doesn't specify the population so I just used n= 1000. This is what I have so far:
A = Someone in the population with Hepatitis (.007)
A' = Someone in the population without Hepatitis (.993)
B = Event that diagnostic test is positive when they have Hep, I'm not sure about this, but I got
.007 * .98 = .00686.
B' = Event that diagnostic test is negative when they have Hep, Unsure with this one too:
.993 * .98 = .97314
Now, P(A) = .007*1000 = 7
P'(A) = 1000 - 7 = 993
P(B|A) = 7 * .00686 = 04802
P(B|A') = 993 * .97314 = 966.32
I'm trying to find: P(A|B) = given that the test turned out positive, the probability that the person has the disease.
So would it be: P(A and B) / P(A) = 4.802^-5 / 7 = 6.86^-6.
Actually, that's wrong cause it's not one of my answer choices. Can you tell me what I did wrong. Or would it be easier to tell me the few things I did right?
for any event A
0<=P(A)<=1
7 or 933 or ... are not probabilities
That's what I initially thought, but then I looked at the example problem labeld "Bayes Theorem / Contingency Table" there are probabilities on there that are greater than 1...or maybe I'm mistaken. But then how do I find
P(T|H) and P(T|H'). Following the equation you gave,
P(H|T)=P(T|H)*P(H)/[P(T|H)*P(H)+P(T|H')*P(H')]
=P(T|H)*.007/[P(T|H)*.007+P(T|H')*P.993]
Hopefully I got the right values for P(H) and P(H'). As for P(T|H) and P(T|H') though, I need help calculating. Thank you
Oh I almost forgot to ask, would you consider a z-score of ~ 2 to be a big deviation from the mean?
P(H)=.005...
Hopefully I got the right values for P(H) and P(H'). As for P(T|H) and P(T|H') though, I need help calculating. Thank you
P(H')=1-.005
P(T|H)=.98
P(T|H')=1-P(T'|H')=1-.98
There's a typo in my original post. It's supposed to be .007 as opposed to
.005. That means it should be,
(.98) * .007 / (.98 * ,007) + (.02 * .998) = (.98)(.26197) = .2567
That seems somewhat right....
Can you get me some tips on the golf problem too? Thanks for all your help.
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