Similar problems: Probability of woman receiving positive mammogram and having cancer Probability problem on a cancer test.

I can't figure this out, and it's really frustrating. The problem is almost identical to the ones linked above, but with different values and a different given.

Given: Probability of being diagnosed with breast cancer is 2.38% (See note)
Given: False-negative rate is 15%
Given: False-positive rate is 5%
Find: Probability of having breast cancer given a positive diagnosis (See note)

Note: The exact wording of the first given is "For women between 50 and 59, there is a 2.38% chance of being diagnosed with breast cancer." The exact wording of the question is "If a woman over 50 has a mammogram and it comes back positive for breast cancer, what is the probability that she has the disease?". Since we aren't given the rate of diagnosis for all women over 50, I'm inclined to assume that either the problem as stated is impossible to solve (more on that at the very end), or that we're meant to understand that the rate of diagnosis for all women over 50 is the same as the the rate for women between 50 and 59, or that "a woman over 50" is meant to imply "a woman between 50 and 59". In the interest of attempting to solve the problem I assumed the latter, as it seems like the most likely intent.

To reiterate ("+": a positive diagnosis; "-": a negative diagnosis; "C": presence of breast cancer; "C'": absence of breast cancer):
P(+) = .0238
P(-|C) = .15
P(+|C) = .85
P(+|C') = .05
P(-|C') = .95
P(C|+) = ???

Using Bayes's Theorem, we can see that P(C|+) = [P(+|C) * P(C)]/P(+)

P(C|+) = .85P(C)/.0238

But we don't have P(C). Using the law of total probability, we then have

P(C) = P(C|+)*P(+)+P(C|-)*P(-)
Well, we can use Bayes's Theorem again to get that
P(C|-) = .95P(C)/.9762
Which allows us to substitute what we have into the law of total probability as

P(C) = .85P(C) * .0238/.0238 + .95P(C) * .9762/.9762
P(C) = .85P(C) + .95P(C)
P(C) = 1.8P(C)
And I think we can all agree that that just doesn't make sense. A value can't be equal to a non-identity multiple of itself.

So I tried using the law of total probability the other way.
P(+) = P(+|C) * P(C) + P(+|C') * P(C')
Since we don't know P(C'), I substituted (1-P(C)) for it to reduce the number of unknown variables.
P(+) = .85P(C) + .05(1-P(C)) = .85P(C) - .05P(C) + .05 = .8P(C) + .05
and since we know the value of P(+)...
.0238 = .8P(C) + .05
.8P(C) = -.0262
P(C) = -.03275
That can't possibly be correct either.

Combining Bayes's and the law of total probability is equally unhelpful.
P(+|C) = P(C|+)P(+)/[P(C|+)P(+) + P(C|-)P(-)]
This would be easily simplified if we knew the probability of cancer given a negative diagnosis.
P(C|-) = P(-|C)P(C)/P(-) = P(-|C)P(C)/[P(-|C)P(C) + P(-|C')P(C')]
Filling in what we know allows us to solve for P(C)
.15P(C)/.0238 = .15P(C)/.95 - .8P(C)
.0238 = .95 - .8P(C)
-.9262 = -.8P(C)
P(C) = 1.15775
And unless my teacher is claiming that every woman is guaranteed to get breast cancer at least once in her life, I sincerely doubt the probability of this outcome is supposed to be greater than 1.

Since every outcome I can reach for P(C) is either negative, greater than 1, or a multiple of itself, I can't find the solution.

Please tell me what I'm doing wrong. As far as I can tell, the only difference between this problem and those I linked is that the linked problems were given the rate of breast cancer rather than the rate of diagnosis of breast cancer. I don't understand why that would make the entire problem unsolvable, but I can't piece together a solution


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.