Question

HIV testing Enzyme immunoassay (EIA) tests are used to screen blood specimens for the presence of antibodies to HIV, the virus that causes AIDS. Antibodies indicate the presence of the virus. The test is quite accurate but is not always correct. A false positive occurs when the test gives a positive result but no HIV antibodies are actually present in the blood. A false negative occurs when the test gives a negative result but HIV antibodies are present in the blood. Here are approximate probabilities of positive and negative EIA outcomes when the blood tested does and does not actually contain antibodies to HIV: 23

Suppose that 1% of a large population carries antibodies to HIV in their blood. Imagine choosing a person from this population at random. If the person’s EIA test is positive, what’s the probability that the person has the HIV antibody?

Short answer

$\text{Probability that the randomly selected person with positive EIA test has the HIV antibody is approx.}0.6270.$

Step-by-step solution

Step 1:Given information

Step 2;Calculation

According to complement rule,

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

According to general multiplication rule,

$P(\mathrm{A}\text{and}\mathrm{B})=P(\mathrm{A}\cap \mathrm{B})=P(\mathrm{A})\times P(\mathrm{B}\mid \mathrm{A})=P(\mathrm{B})\times P(\mathrm{A}\mid \mathrm{B})$

According to addition rule for mutually exclusive event,

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

Conditional probability definition:

$P(\mathrm{B}\mid \mathrm{A})=\frac{P(\mathrm{A}\cap \mathrm{B})}{P(\mathrm{A})}=\frac{P(\mathrm{A}\text{and}\mathrm{B})}{P(\mathrm{A})}$

Let

A: Antibodies present

$A^c$: Antibodies absent

P: Positive ElA test

Pc: Negative EIA test

Now,

The corresponding probabilities:

Probability for Antibodies present,

$P\left(A\right)=0.01$

Probability for Antibodies present Positive EIA test,

$P(\mathrm{P}\mid \mathrm{A})=0.9985$

Probability for Antibodies present Negative EIA test,

$P\left({\mathrm{P}}^{\mathrm{c}}\mid \mathrm{A}\right)=0.0015$

Probability for Antibodies absent Positive EIA test,

$P\left(\mathrm{P}\mid {\mathrm{A}}^{\mathrm{c}}\right)=0.006$

Probability for Antibodies absent Negative ElA test,

$P\left({\mathrm{P}}^{\mathrm{c}}\mid {\mathrm{A}}^{\mathrm{c}}\right)=0.994$

Apply complement rule:

Probability for antibodies absent,

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

Apply general multiplication rule:

Probability for Positive EIA test and Antibodies present,

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

Probability for Positive EIA test and Antibodies absent,

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

Since the Antibodies are either absent or present,

Apply general addition rule for mutually exclusive events:

Probability for positive EIA test,

$P\left(P\right)=P\left(P\text{and}\mathrm{A}\right)+P\left(\mathrm{P}\text{and}{\mathrm{A}}^{\mathrm{c}}\right)$

$=0.009985+0.00594$

$=0.015925$

Using conditional probability definition:

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

Thus,

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