False positive and false negative probability question?

Approximately 25% of Canadian men in the 50's have prostate cancer. High level of PSA in the blood is used as a test for prostate cancer. Approximately 20% of men with prostate cancer have normal PSA levels and two out of three men without prostate cancer have high PSA levels. Calculate the probabilities that a PSA test on a Canadian man in his 50's will result in


a) false positive


b) false negative





Thanks for the help :)False positive and false negative probability question?
P(had) = 25% = 1/4


P(not) = 3/4


P(normallhad) = 20% = 1/5


P(highlnot) = 2/3





qa


P(false positive)


= P(positive but not) / P(positive)


= P(highlnot)*P(not) / [P(highlnot)*P(not) + P(highlhad)*P(had)]


= 2/3 * 3/4 / [2/3 * 3/4 + 4/5 * 1/4]


= 1/2 / [1/2 + 1/5]


= 5/7





qb


P(false negative)


= P(negative but had) / P(negative)


= 1/5 * 1/4 / [1/5 * 1/4 + 1/3 * 3/4]


= 1/20 / [1/20 + 1/4]


= 1/6False positive and false negative probability question?
This is a conditional probability question, or Boyles Theorem:


You will need to find all intersections, these are elementary events that add up to 1.





Let n represent normal PSA level.


P(c) = .25 so P(∼c) = .75


P(n⎮c)=.20


P(n⎮∼c) = 2/3


Now that is what is given. You what to know given the test is not normal what is the chance a man does not have cancer, or if the test is normal what is the chance he does have cancer.


a) P( ∼c⎮∼n)


b) P(c⎮n)


To do thse problems you need to find the intersections first. Sketch two intersecting circles one for n and one for c this will divide the population into the four intersections youwill need.


P(c∩n) and P(∼c∩∼n) first


Keep in mind conditional prob definition P(n⎮c) = P(c∩n)/P(c)


P(c∩n) = P(n⎮c)*P(c) = .20*.25 = .05


P(∼c∩∼n) = P(∼n⎮∼c)*P(∼c) = 2/3*.75 = .5


To find the P(n): P(n) = P(n∩c)+P(n∩∼c)


This are the sum of two intersections, elementary events, You all ready know two (above) and P(c) = .25


since


P(c) = P(c∩n) + P(c∩∼n),


P(c∩∼n) = .20


You now know three elementary events subtracting the three intersections you know from 1.


P(∼c∩n)=.25


P(n) = P(n∩c) + P(n∩∼c)


a) P( ∼c⎮∼n) = P(∼c∩∼n)/P(∼n) =P(∼c∩∼n)/(P(∼c∩∼n)+P(c∩∼n)) =.5/(.5+.2)=5/7≈.714


b) P(c⎮n)= P(c∩n)/P(n)= P(c∩n)/(P(c∩n)+P(∼c∩n))=


.05/(.05+.25)= .05/.3 ≈ .167