# Probability of woman receiving positive mammogram and having cancer

The probability that a randomly selected US woman will have breast cancer in their lifetime is 0.12. Women over 40 are advised to have regular mammograms because early detection of breast cancer means treatment can be started earlier and with higher expectations for survival. Mammograms are good, but not definitive in a cancer diagnosis. One way to express the precision of mammography is

P(Positive Mammogram | Cancer ) = 0.78

P(Negative Mammogram | No Cancer ) = 0.90

If a woman receives a positive mammogram (that is, the radiologist advises meeting with an oncologist), what is the probability that she has breast cancer? That is, compute

P(Cancer | Positive Mammogram ).

I'm pretty sure the info at the beginning is irrelevant. I can't figure out what information I need to calculate. I've come up with 4 numbers, and I think I need to multiply 2 of them to get my answer:

P(B│A)=0.78 prob. cancer=true, mammogram=true

P(Bc│Ac)=0.90 prob. cancer=false, mammogram=false

P(B|Ac)=1-0.78=0.22 prob. cancer=false, mammogram=true

P(Bc|A)=1-0.90=0.10 prob. cancer=true, mammogram=false

• We want $\Pr(C|P)$ which is $\frac{\Pr(C\cap P)}{\Pr(P)}$. In my experience fewer students get confused if they use this than if they use the Bayes' Rule version. – André Nicolas Jan 29 '15 at 7:03
• @AndréNicolas I think that the OP is exchanging the two sets in the definition of $P(A|B)$. – Emanuele Paolini Jan 29 '15 at 7:14
• @EmanuelePaolini: It is a risk when here are letters and complemented letters flying all over. That's why your geometric answer is best, or possibly two answers given together, the first geometric and the second using the definition of conditional probability. After the picture, the symbols $\Pr(P)$, $\Pr(C\cap P)$ begin to make sense. – André Nicolas Jan 29 '15 at 7:25
• @AndréNicolas Thanks for the tips--I think I'll need a bit more hand-holding, as I'm just starting. So Pr(P) represents area (B) in the picture and Pr(C∩P) is represented by the area of intersection of A&B in the picture. So we need to figure out Pr(C∩P). Of those who have a positive mammogram, 0.78 have cancer. Do we need to implement the 0.12 at this point? As for the probability of a positive test, Pr(P), we need to find (1)those who have a positive test AND have cancer and (2)those who have a positive test and DON'T have cancer...I'm not sure how to get the latter. (1-0.90?) – wad11656 Jan 29 '15 at 21:27
• $\Pr(C\cap P)=(0.12)(0.78)$. And $\Pr(P)$ is $(0.12)(0.78)+(0.88)(0.10)$. This is because there are two ways one can test positive, have the disease and get a positive or not have the disease and get a (false) positive on the test. – André Nicolas Jan 29 '15 at 21:42

Hint:

The information at the beginning is not irrelevant.

Suppose there are 1 million women tested.

• How many might have breast cancer and how many not?
• How many might have cancer and a positive test?
• How many might not have cancer and have a negative test?
• How many might not have cancer but have a positive test?
• How many might have a positive test?

What proportion of those with a positive test might have cancer?

You can represent this kind of problem as in the picture below. The region $A$ (red) represents women with breast cancer. The region $B$ represents women with positive mammogram. If the square has total area 1, the area of $A$ is 0.12, which is very important to know. Conditional probability, like for example $P(B|A)$ is the area of the intersection $A\cap B$ with respect to (knowing that) the area of $B$.

So

The probability that a randomly selected US woman will have breast cancer in their lifetime is 0.12

This first statement is to give us an estimate of the baseline cancer rate. The wording "randomly selected US woman" is intended to imply that if we repeated this process an infinite number of times (i.e., sampling with replacement), 12% of our encountered subjects would experience cancer at some point in their lives.

$$P(\textrm{Cancer}) = P(C) = 0.12$$ - although see note below

Next we are given information about testing accuracy:

$$P(+| C ) = 0.78$$ - Probability of a positive test given you do have cancer

$$P(- | NC ) = 0.90$$ - Probability of a negative test given you do not have cancer (a double negative)

With this information we can start to solve the quantity of interest, the probability of cancer given a positive diagnosis (i.e., $$P(C|+)$$), which can be solved using Bayes' theorem as

$$P(C|+) = {P(+|C) \cdot P(C) \over P(+)}$$

We know $$P(+|C)$$ and $$P(C)$$, but we do not know the overall probability of a positive diagnosis, $$P(+)$$. This is straightforward to determined as the population can be divided into two non-overlapping populations, those with cancer and those without cancer - you either have it or you don't. If we know the positive diagnosis rates for both populations, we can determine the overall diagnosis probability as:

$$P(+) = P(+|C)\cdot P(C) + P(+|NC) \cdot P(NC)$$

We are given $$P(+|C)$$, the positive diagnosis rates for patients with cancer, but do not directly know $$P(+|NC)$$ the false positive rate. Assuming the test results come back either positive or negative (i.e., there is not a third option of N/A or unknown), we can first determine the false positive rate as the following compliment:

$$P(+|NC) = 1 - P(-|NC) = 1 - 0.9 = 0.1$$

Next, the probability of not having cancer, $$P(NC)$$, can be determined in a similar manner because subjects can only belong to one of the two populations, with or without cancer (i.e., $$P(NC) = 1 - 0.12)$$). This lets us solve the probability of a positive diagnosis across the entire population as: $$P(+) = 0.78 \cdot 0.12 + 0.1 \cdot (1- 0.12) = 0.1816$$ Note that, this is of course higher than the actual cancer rate due to the relatively high false positive rate of the test.

Finally, we can solve our original Bayes' equation as
$$P(C|+) = {0.78\cdot 0.12 \over 0.1816} \sim 0.515$$

Note: where this question is poorly worded is that we don't know the instantaneous probability that a subject has cancer, we are only given the lifetime risk. For the hypothetical patient it will be lower than 0.12 as you develop cancer at some point in your life, you do not have cancer over the duration of your life (or necessarily the ages in which you would be exposed to testing), therefore all we can really say is that $$P(C) < 0.12$$. We can however visualize the diagnosis rages changes for different baseline rates.

The intercept goes through zero as the probability the patient has cancer approach zero as the prevalence of cancer in the population goes to zero.