Question involving Bayes' Rule

Question

The question is as follows:

Suppose a screening test for AIDS has the following features: (i) If a blood sample comes from someone with AIDS then the test will be positive 95% of the time. (ii) If the blood sample comes from someone without AIDS then the test will be negative 95% of the time. Suppose that 5% of the population has AIDS. If a blood sample tests positive, what is the probability that the person whose blood was tested has AIDS?

Now I think the approach to solve this comes from Bayes' Rule. The partition $H_i$ would be if the person has AIDS or not, and the event $A$ would be if the test is positive or not. Would I be correct in assuming this?

Bayes' Rule:

$$\mathbb{P}(H_i\mid A) = \frac{\mathbb{P}(H_i) \mathbb{P}(A\mid H_i)}{\sum_{j=0}^\infty\mathbb{P}(H_j)\mathbb{P}(A\mid H_j)}$$

My guess is :

$$\mathbb{P}(H_i\mid A) = \frac{(0.05)(0.95)}{(0.05)(0.95) + (0.95)(*)(0.95)^2}$$

The star is where I'm not 100% sure.

Answer 1

The chance of someone testing positive while having AIDS (=0.95*0.05) is the same as the chance of someone testing positive who does not have AIDS (=0.05*0.95)!

So, if you test positive, you are equally likely to have AIDS as to not have AIDS. So, if a person under these conditions tests positive, they have a 50% chance of having AIDS.

Answer 2

Here's one way to do it when the partition of the sample space into two hypotheses ("AIDS" and "no AIDS") has only two parts: $$ \frac{\Pr(\text{AIDS} \mid \text{positive})}{\Pr(\text{no AIDS}\mid\text{positive})} = \frac{\Pr(\text{AIDS})}{\Pr(\text{no AIDS})} \times \frac{\Pr(\text{positive}\mid\text{AIDS})}{\Pr(\text{positive} \mid \text{no AIDS})} = \frac 5 {95} \times \frac{95} 5 = 1. $$ Thus $\Pr(\text{AIDS}\mid\text{positive}) = \Pr(\text{no AIDS}\mid \text{positive}).$

Answer 3

Let us say that $H_i \in \{\text{aids}, \text{healthy}\}$ and $A \in \{\text{positive},\text{ negative}\}$. Next, from the problem statement we know that

\begin{align} &\mathbb{P}(A = \text{positive} \mid H_i = \text{aids}) = 0.95\\ &\mathbb{P}(A = \text{negative} \mid H_i = \text{healthy}) = 0.95 \\ & \mathbb{P}(H_i = \text{aids}) = 0.05 \tag 1 \end{align}

and we are asked for the value of $\mathbb{P}(H_i = \text{aids} \mid A = \text{positive})$. Exactly as you proposed, Bayes' rule is what has to be used here. In particular we use

\begin{equation*} \mathbb{P}(H_i = \text{aids} \mid A = \text{positive})= \frac{\mathbb{P}(A = \text{positive} \mid H_i = \text{aids}) \mathbb{P}(H_i = \text{aids}) }{\mathbb{P}(A = \text{positive})} \tag 2 \end{equation*}

From (1) we have all values in the left part of (2) except for the evidence $\mathbb{P}(A = \text{positive})$, which can be computed marginalizing over $H_i$ in the joint probability $\mathbb{P}(A = \text{positive},H_i)$, as highlighted in (3).

\begin{align} \mathbb{P}(A = \text{positive}) = {} & \sum_{H_i} \mathbb{P}(A = \text{positive},H_i) = \sum_{H_i} \mathbb{P}(A = \text{positive}\mid H_i) \mathbb{P}(H_i)\\ = {} & \mathbb{P}(A = \text{positive}\mid H_i = \text{aids}) \mathbb{P}(H_i = \text{aids}) \\ & {} + \mathbb{P}(A = \text{positive}\mid H_i = \text{healthy}) \mathbb{P}(H_i = \text{healthy}) \tag 3 \end{align}

Note that in (3) two new probabilities appear, which however can easily be computed using the complementary probabilities of the given data in (1) as

\begin{equation} \begin{split} &\mathbb{P}(A = \text{positive} \mid H_i = \text{healthy}) = 1-\mathbb{P}(A = \text{negative} \mid H_i = \text{healthy}) = 0.05\\ & \mathbb{P}(H_i = \text{healthy}) = 1- \mathbb{P}(H_i = \text{aids}) = 0.95 \end{split} \tag 4 \end{equation}

Substituting the results from (4) in (3) we obtain that $\mathbb{P}(A = \text{positive}) = 0.095$. Finally, using this results we compute (2) as

\begin{equation*} \mathbb{P}(H_i = \text{aids} \mid A = \text{positive})= \frac{0.95\cdot0.05}{0.095} = 0.5 \end{equation*}

The test turns out to be not really trustful. This is basically due to high population percentage having aids.

P.S. I recommend you the reading of these two articles, they are very interesting and can give you rich insights.

• Prosecutor's Fallacy
• Base Rate Fallacy