How Accurate Is the New Blood Test in Detecting Disease?

[lunds002]

A new blood test has been shown to be effective in the early detection of a disease. The probability that the blood test correctly identifies someone with this disease is 0.99, and the probability that the blood test correctly identifies someone without that disease is 0.95. The incidence of this disease in the general population is 0.0001.

A doctor administered the blood test to a patient and the test result indicated that this patient had the disease. What is the probability that the patient has the disease?

 

[micromass]

Please post an attempt...

 

[lunds002]

(0.0001 x 0.99) / ((0.0001 x 0.99)+(.0001 x 0.95)) = 0.051

Here I did the probability of favorable over total, but I'm not sure this is correctly done.

 

[micromass]

(0.0001 x 0.99) / ((0.0001 x 0.99)+(.0001 x 0.95)) = 0.051

Here I did the probability of favorable over total, but I'm not sure this is correctly done.

Hmmm, I'm a bit skeptical about the (.0001 x 0.95) part. But it could be that I misinterpret the question.

Let's put some notation in. Denote
A=event that the person has the disease
B=event that the test is positive (i.e. says that the person has the disease)

When the question says

the probability that the blood test correctly identifies someone without that disease is 0.95.

It does mean "if the person does not have the disease, then the test indicates that the person does not have the disease", right?

So we have

$$P(B^c|A^c)=0.95$$

Now, you applied Bayes' law, that says

$$P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)}$$

You claim that the term $P(B|A^c)P(A^c)$ corresponds to (.0001 x 0.95), but this is not correct... Try to calculate the real probabilities.

 

[lunds002]

Yes that was the correct interpretation.

Okay, so maybe this makes more sense:

(.0001 x .99) / ((.0001 x .99) + (.95 x .01)) = 0.0103

 

[micromass]

Not really correct...

What is

$$P(A^c)~\text{and}~P(B|A^c)$$

and how did you calculate them?

 

[lunds002]

(.0001 x .99) / ((.0001 x .99) + (.01 x .999)) = 0.009

Okay I don't think this is right either, but I got the (.01 x .999) by taking 1 - P(correctly identifies with disease) to get P(incorrectly identifies with disease)=.01. Then I took 1 - P(has disease) to get P(don't have disease)=.999

 

[micromass]

(.0001 x .99) / ((.0001 x .99) + (.01 x .999)) = 0.009

Okay I don't think this is right either, but I got the (.01 x .999) by taking 1 - P(correctly identifies with disease) to get P(incorrectly identifies with disease)=.01.

No, but you have the right idea. In fact, you need to do

$$P(B|A^c)=1-P(B^c|A^c)$$

Do what you want of course, but I highly suggest using the mathematical notation for things. This will eliminate your chance on mistakes a lot! In fact, I think this is what went wrong here.

Then I took 1 - P(has disease) to get P(don't have disease)=.999

Correct, but it needs to .9999

 

[lunds002]

Okay I'll give it another shot.

A=event that the person has the disease
B=event that the test is positive (i.e. says that the person has the disease)

so P(A|B) = P(A) x P(B|A)
P(A) x P(B|A) + P(A') x P(B|A')

P(A) = .0001
P(B|A) = .99
P(A') = 1 - P(A) = .9999
P(B|A') = 1 - P(B'|A') = 1 - (.95 x .9999) = .05 <-- not sure here
So then P(A|B) = .0001 x .99 / (.0001 x .99 + .9999 x .05) = 0.00198 This number should be larger, i would think.

 

[micromass]

Okay I'll give it another shot.

A=event that the person has the disease
B=event that the test is positive (i.e. says that the person has the disease)

so P(A|B) = P(A) x P(B|A)
P(A) x P(B|A) + P(A') x P(B|A')

P(A) = .0001
P(B|A) = .99
P(A') = 1 - P(A) = .9999
P(B|A') = 1 - P(B'|A') = 1 - (.95 x .9999) = .05 <-- not sure here

Why do x.9999? You have P(B'|A')=0.95, so there's no need for multiplying. But
P(B|A')=0.05 is correct, though.

So then P(A|B) = .0001 x .99 / (.0001 x .99 + .9999 x .05) = 0.00198 This number should be larger, i would think.

This is correct. And it is indeed quite surprising that the number is so low! The thing is that the disease is so rare that it is more likely that the test was incorrect than that the person actually has the disease!

 

[lunds002]

Wait so my answer was correct?

 

[micromass]

Yes!

 

[lunds002]

Great!
Thanks so much for helping me through that, it was a bit of a struggle for me.

 

[Ray Vickson]

Okay I'll give it another shot.

A=event that the person has the disease
B=event that the test is positive (i.e. says that the person has the disease)

so P(A|B) = P(A) x P(B|A)
P(A) x P(B|A) + P(A') x P(B|A')

P(A) = .0001
P(B|A) = .99
P(A') = 1 - P(A) = .9999
P(B|A') = 1 - P(B'|A') = 1 - (.95 x .9999) = .05 <-- not sure here
So then P(A|B) = .0001 x .99 / (.0001 x .99 + .9999 x .05) = 0.00198 This number should be larger, i would think.

Another way to think about it which you may, or may not find more intuitive (some people do, others don't) is: imagine giving the test to a population of, say 1,000,000 people. Of these, there are 0.0001*1000000 = 100 who have the disease, while 1000000-100 = 999900 do not have it. Among the 100 having the disease, 99 will give a positive test result. Among the 999900 not having the disease, 5% of them will test positive, so the number positive will be 0.05*999900 = 49995. Therefore, in the whole population, the number testing positive is Npos = 499995 + 99 = 50094 (so P{test pos} = 50094/1000000 = 0.050094). Given a positive test, the probab. the person actually has the disease is 99/50094 = 1/506 = 0.001976 = P{disease|test pos}.

Of course, the Bayesian formulas give the same result exactly.

RGV