Chapter 5.3, Problem 88E

To determine

Probability for a person getting positive result on both EIA tests has HIV antibodies.

Answer to Problem 88E

Probability that a person gets positive result on both EIA tests has HIV antibodies is approx. 0.3931.

Explanation of Solution

Given information:

Probability for Antibodies present,

  $P\left(\text{A}\right)=1\%=0.01$

Probability for Antibodies present Positive EIA test,

  $P\left(\text{P}|\text{A}\right)=0.9985$

Probability for Antibodies present Negative EIA test,

  $P\left({\text{P}}^{\text{c}}|\text{A}\right)=0.0015$

Probability for Antibodies absent Positive EIA test,

  $P\left(\text{P}|{\text{A}}^{\text{c}}\right)=0.0060$

Probability for Antibodies absent Negative EIA test,

  $P\left({\text{P}}^{\text{c}}|{\text{A}}^{\text{c}}\right)=0.9940$

Calculations:

According to the multiplication rule for independent events,

  $P\left(\text{A}\cap \text{B}\right)=P\left(\text{A}\text{and}\text{}\text{B}\right)=P\left(\text{A}\right)\times P\left(\text{B}\right)$

Conditional probability definition:

  $P\left(\text{B}|\text{A}\right)=\frac{P\left(\text{A}\cap \text{B}\right)}{P\left(\text{A}\right)}=\frac{P\left(\text{A}\text{and}\text{B}\right)}{P\left(\text{A}\right)}$

Let

B: Positive result on both EIA tests

P: Positive result on one EIA test

A: Antibodies present

From the previous problem,

We have

Probability for positive EIA test,

  $P\left(P\right)=0.015925$

And

Probability for Positive EIA test and Antibodies present,

  $P\left(\text{P}\text{and}\text{A}\right)=P\left(\text{A}\right)\times P\left(\text{P}|\text{A}\right)=0.01\times 0.9985=0.009985$

Now,

Apply multiplication rule for independent events, to get theprobability for positive result on both EIA tests,

  $\begin{array}{l}P\left(\text{B}\right)=P\left(\text{P}\right)\times P\left(\text{P}\right)\\ =0.015925\times 0.015925\\ =0.000253605625\end{array}$

Also,

Apply multiplication rule for independent events, to get the probability for antibodies present and positive result on both EIA tests,

  $\begin{array}{l}P\left(\text{A}\text{and}\text{B}\right)=P\left(\text{A}\text{and}\text{P}\right)\times P\left(\text{A}\text{and}\text{P}\right)\\ =0.009985\times 0.009985\\ =0.00009970225\end{array}$

Using conditional probability definition:

  $P\left(\text{A}|\text{B}\right)=\frac{P\left(\text{A}\text{and}\text{B}\right)}{P\left(\text{B}\right)}=\frac{0.000099700225}{0.000253605625}\approx 0.3931$

Thus,

The conditional probability for a person getting positive result on both EIA tests has HIV antibodies is approx. 0.3931.