A small probability correction
September 29, 2026 at 10:35 AM by Dr. Drang
This short video appeared in my YouTube feed last week. It’s from Hannah Fry, whom you probably know from her appearances on Numberphile and other STEM-oriented stuff. If you’re in the UK, you’ve may have seen her on the BBC, too.
The short presents a classic problem in conditional probability, but the answer she comes up with is wrong. It’s a good estimate, and it’s the same answer I got when I stopped the video and tried to work it out in my head, but it’s still wrong. By just a little bit.
Here’s the problem: There’s a disease that affects 1 in 1,000 people. A test for the disease is perfect in one sense but imperfect in another. If you have the disease, the test will return a positive result 100% of the time. If you don’t have the disease, the test will return a negative result 95% of the time but a positive result 5% of the time. If you have the test and the result is positive, what is the probability you have the disease?
What makes this a classic problem is that it presents you with conditional probability in one sense (the probability of a positive test given that you have the disease) and asks for a conditional probability in the opposite sense (the probability that you have the disease given that your test was positive). The solution combines the definition of conditional probability, the commutative property of intersections, and the law of total probability.
Let’s define some events. is you having the disease, and is you getting a positive test result. Putting a horizontal bar over these indicates not having the disease and not testing positive (i.e., testing negative), respectively. Therefore
The vertical bars are read as “given,” meaning the event after the bar is the condition. What we’ve been asked to find is . Let’s work it out.
By the definition of conditional probability, we can say
where means the intersection of the two events. So
Because the intersection of events is commutative
Both terms on the right-hand side of this equation are known, so we can say
Since and are mutually exclusive and collectively exhaustive, the law of total probability says
So
which is, as I said, pretty close to the 2% answer in the video but not exactly.
This formal approach is how you’re taught to solve problems like this in an introductory probability class, but Dr. Fry and I used a more concrete method to get our nearly correct answers. Here’s what we did:
Imagine 1,000 typical people. Of these, 1 should have the disease (correct) and 50 should test positive (incorrect). That tells us that 1 in 50, or 2%, of the people who test positive will have the disease. Here’s a screenshot from the video that matches this calculation:
What makes this calculation wrong is that it implicitly assumes that 5% of everyone will test positive, not 5% of only those who don’t have the disease. The number who will test positive should be 5% of 999, which is 49.95, plus the 1 who does have the disease. So 1 in 50.95, or 1.9627%, of those who test positive will have the disease. This answer matches that of the formal approach.
This is somewhat unsatisfying, though, as the purpose of this “imagine a bunch of typical people” method is to have all the people counts be integers. Although the numbers work out when you get to the end, it’s distracting to litter the discussion with fractional people. You can get around this by imagining more people—a million, say—but then all the numbers get bigger: 1,000 people with the disease and 49,950 false positives. This isn’t a problem for the kind of people who read this blog but isn’t so great for the more general audience Dr. Fry is addressing.
Personally, I would have been OK with her saying that 5% of 999 is almost 50, so the number who test positive is nearly 51. And 1 out of 51 is just under 2%—call it 2% in round figures. You still make the point that it’s way less than 95% and that problems like this require some care.