Chapter 5.3, Problem 85E

To determine

Probability for the randomly selected person with positive EIA test has the HIV antibody.

Answer to Problem 85E

Probability that the randomly selected person with positive EIA test has the HIV antibody is approx. 0.6270.

Explanation of Solution

Given information:

Data for African and Asian ancestry in United States:

82% of the population is White

14% of the population is Black

4% of the population is Asian

Lactose intolerant people:

15% are Whites

70% are Blacks

90% are Asians

Calculations:

According to complement rule,

  $P\left({\text{A}}^{\text{c}}\right)=P\left(\text{not}\text{A}\right)=1-P\left(\text{A}\right)$

According to general multiplication rule,

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

According to addition rule for mutually exclusive event,

  $P\left(\text{A}\cup \text{B}\right)=P\left(\text{A}\text{or}\text{}\text{B}\right)=P\left(\text{A}\right)+P\left(\text{B}\right)$

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

A: Antibodies present

A^c: Antibodies absent

P: Positive EIA test

P^c: Negative EIA test

Now,

The corresponding probabilities:

Probability for Antibodies present,

  $P\left(\text{A}\right)=0.01$

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,

  $P\left(\text{P}|{\text{A}}^{\text{c}}\right)=0.006$

Probability for Antibodies absent Negative EIA test,

  $P\left({\text{P}}^{\text{c}}|{\text{A}}^{\text{c}}\right)=0.994$

Apply complement rule:

Probability for antibodies absent,

  $P\left({\text{A}}^{\text{c}}\right)=1-P\left(\text{A}\right)=1-0.01=0.99$

Apply general multiplication rule:

Probability for Positive EIA test and Antibodies present,

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

Probability for Positive EIA test and Antibodies absent,

  $P\left(\text{P}\text{and}{\text{A}}^{\text{c}}\right)=P\left({\text{A}}^{\text{c}}\right)\times P\left(\text{P}|{\text{A}}^{\text{c}}\right)=0.99\times 0.006=0.00594$

Since the Antibodies are either absent or present,

Apply general addition rule for mutually exclusive events:

Probability for positive EIA test,

  $\begin{array}{l}P\left(P\right)=P\left(\text{P}\text{and}\text{A}\right)+P\left(\text{P}\text{and}{\text{A}}^{\text{c}}\right)\\ =0.009985+0.00594\\ =0.015925\end{array}$

Using conditional probability definition:

  $P\left(\text{A}|\text{P}\right)=\frac{P\left(\text{P}\text{and}\text{A}\right)}{P\left(\text{P}\right)}=\frac{0.009985}{0.015925}=\frac{9985}{15925}=\frac{1997}{3185}\approx 0.6270$

Thus,

The conditional probability for randomly selected person with positive EIA test has the HIV antibody is approx. 0.6270.