Probability of breast cancer I'm having trouble with a probability problem I've been trying to solve for a while. It's about the accuracy of breast cancer testing. Relevant probabilities are listed below, where:

*

*"$\text{cancer}$" is the event "has breast cancer".

*"$+$" is the event "tests positive for breast cancer".

$P(\text{cancer}) = \frac{12}{1000}$
$P(+|\text{cancer}) = \frac{11}{12}$
$P(+) = \frac{31}{1000}$
$P(\text{cancer}|+) = 0.355$
This last line is a result from a previous problem. The next part involves updating the probability of having cancer, but I'm having trouble figuring out what the answer is.
In the next part of the question, there's a woman who has tested positive and her doctor says she's part of a population for which there's a 40% chance of breast cancer.
I need to find the probability that the woman has cancer.
I am confused by this update to the cancer probability, but i will assume that this means $P(\text{cancer})$ has changed.
I also assume this means I need to find a new value for $P(\text{cancer}|+)$, but I'm not getting this right.
$P(+ | \text{cancer}) = \frac{11}{12} = \frac{P(\text{cancer} | +)\cdot P(+)}{P(\text{cancer})} = \frac{P(\text{cancer} | +) \cdot \frac{31}{1000}}{0.40}$
$P(\text{cancer} | +) = \frac{11}{12} \cdot 0.40 \cdot \frac{1000}{31} = 11.828$
The result can't be correct because it's way over 1.
How can i fix this? Thank you in advance for any insight.
 A: By Bayes' Theorem:
$$P(cancer|+) = \frac{P(+|cancer) P(cancer)}{P(+)}$$
Where $$P(+) = P(+|cancer)P(cancer)+ P(+|no-cancer)P(no-cancer)$$
$$P(+|cancer) = \frac{11}{12}$$
$$P(cancer) = \frac{4}{10}$$
Therefore,
$$P(cancer|+) = {\frac{11}{12} \cdot \frac{4}{10} \over \frac{11}{12} \cdot \frac{4}{10} + \frac{6}{10} \cdot P(+|no-cancer)}$$
To find the rate of false positives for the test, P(+|no-cancer), we can use the information from the general population, that $P(+)= \frac{31}{1000}$ and that $P(cancer) = \frac{12}{1000}$.
Then, $$ \frac{31}{1000} = \frac{11}{12} \cdot \frac{12}{1000} + P(+|no-cancer) \cdot \frac{988}{1000} $$
Rearrange to get $$ P(+|no-cancer) = \frac{5}{247}$$
Plug this back into the previous equation to get $$P(cancer|+) = \frac{2717}{2807} \approx 0.968$$
A: It is natural to assume that in this case the previous value of $P(+)$ is not applicable (it is a very bed test, which has $3.1\%$ chances to be positive in a population where a real chance is $40\%$). Moreover, this contradicts the condition $P(+ | cancer) = \frac{11}{12}$, because then $P(+)$ is at least $\frac{11}{12}\cdot 0.4>0.031$.
It is natural to assume that values of $P(+|cancer)$ and $P(cancer|+)$ reflect efficiency and reliability of the test. I expect that a  testing procedure deals with an isolated sample, so it is independend on cancer spread. But if we keep these values then the probability $P’(cancer|+)$ that the woman has a breast cancer is $P(cancer|+)$, and the information $P’(cancer)=0.4$ is redundant.
So we assume that the testing procedure provides only  $P(+|cancer)$ and $P(+|\neg cancer)$. Then from the given probabilities we have
$$\frac{31}{1000}=P(+)=P(+|cancer)P(cancer)+ P(+|\neg cancer)P(\neg cancer)=$$ $$ \frac{11}{12}\cdot \frac{12}{1000}+ P(+|\neg cancer) \cdot \frac{988}{1000},$$
so $P(+|\neg cancer)=\frac 5{247}$.
Then
$$P’(+)=P(+|cancer)P’(cancer)+ P(+|\neg cancer)P’(\neg cancer)= \frac{11}{12}\cdot 0.4+\frac 5{247}\cdot 0.6=\frac {2807}{7410}.$$
Since $P’(cancer|+) P’(+)=P’(cancer\, \&\, +)= P’(+|cancer) P’(cancer),$
we have
$$P’(cancer|+)=\frac{ P’(+|cancer) P’(cancer)}{P’(+)}=\frac{\frac{11}{12}\cdot 0.4}{\frac {2807}{7410}}=\frac {2717}{2807}\approx 0.968.$$
