1
$\begingroup$

Suppose I have the following events:

  • Event A: Contracted HIV
  • Event B: Tested Positive with HIV

Given $1$ in $1000$ people gets contracted with HIV and the HIV Test gives a wrong test result for $1$ in $100$ people. If a person is tested Positive with HIV, I need to find the probability of him really contracted with HIV.

So I need to find $P(A|B)$ and that $P(A|B)=\frac{P(B|A) \cdot P(A)}{P(B)}$.

To find $P(B)$, I need to find all that are tested positive. So...

$P(B)=\frac{1}{1000} \cdot \frac{99}{100}+\frac{999}{1000} \cdot \frac{1}{100}=\frac{1098}{100000}$

$P(B|A)=\frac{1}{1000} \cdot \frac{99}{100}=\frac{99}{100000}$

$P(A)=\frac{1}{1000}$

Therefore, $P(A|B)=\frac{P(B|A) \cdot P(A)}{P(B)}=\frac{\frac{99}{100000}\cdot \frac{1}{1000}}{\frac{1098}{100000}}=\frac{11}{122000}=0.0000901639=0.009016\%$

While this may seem ok to me at first, I suspected my answer to be wrong because $0.009016\%$ as a figure is a little hard to believe. I did a search on the internet and looks like the answer could be just $ \frac{99}{1098} = 9.016\% $.

Is my initial answer wrong? What has gone wrong with my initial answer of $0.009016\%$? I simply followed the Bayes' Conditional formula.

  • 0
    @xEnOn Yes, that's correct.2012-01-17

1 Answers 1

4

For convenience, let's say that there are 100,000 people (why I chose 100,000 will hopefully be clear shortly). Since you said 1 in 1000 people has HIV, of our 100,000 people, 100 have HIV and 99,900 do not. Since the HIV test returns an incorrect result 1 in 100 times: of the 100 HIV-positive people, 99 will test positive and 1 will test negative; of the 99,900 HIV-negative people, 98,901 will test negative and 999 will test positive. Making a chart:

                  HIV-positive      HIV-negative      total tests positive           99               999         1,098 tests negative            1            98,901        98,902          total          100            99,900       100,000 

Now, you wanted the probability that a person who tests positive is actually HIV-positive. 1,098 people tested positive and of those, only 99 were HIV-positive. $\frac{99}{1098}=\frac{11}{122}\approx 9.01639\%$

As stated in comments, it's your $P(B|A)$ that's off. $P(B|A)$ is the probability that an HIV-positive person tests positive—in the table above, there are 100 HIV-positive people and 99 test positive, so $P(B|A)=\frac{99}{100}.$

You might consider an alternate formula for the conditional probability: $P(A|B)=\frac{P(A\text{ and }B)}{P(B)}.$ $P(A\text{ and }B)$ is the probability of a person testing positive and being HIV-positive, which is the 99 out of 100,000 that you computed, so $P(A|B)=\frac{P(A\text{ and }B)}{P(B)}=\frac{\frac{99}{100000}}{\frac{1098}{100000}}=\frac{99}{1098}=\frac{11}{122}\approx 9.01639\%.$