91Ó°ÊÓ

Probability for Antibodies present,

$P(\mathrm{A})=1\%=0.01$

Probability for Antibodies present Positive ElA test,

$P(\mathrm{P}âˆ\pounds \mathrm{A})=0.9985$

Probability for Antibodies present Negative EIA test,

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

Probability for Antibodies absent Positive EIA test,

$P\left(\mathrm{P}âˆ\pounds {\mathrm{A}}^{\mathrm{c}}\right)=0.0060$

Probability for Antibodies absent Negative ElA test,

$P\left({\mathrm{P}}^{\mathrm{c}}âˆ\pounds {\mathrm{A}}^{\mathrm{c}}\right)=0.9940$

Let

B: Positive result on both EIA tests

P: Positive result on one ElA 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 ElA test and Antibodies present,

$P\left(\mathrm{P}\text{and}\mathrm{A}\right)=P(\mathrm{A})×P(\mathrm{P}âˆ\pounds \mathrm{A})=0.01×0.9985=0.009985$

Now,

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

$P(\mathrm{B})=P\left(\mathrm{P}\right)×P\left(\mathrm{P}\right)$

$=0.009985×0.009985$

$=0.000253605625$

Also,

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

$P(\mathrm{A}\text{and}\mathrm{B})=P(\mathrm{A}\text{and}\mathrm{P})×P(\mathrm{A}\text{and}\mathrm{P})$

$=0.009985×0.009985$

$=0.00009970225$

Using conditional probability definition:

$P(\mathrm{A}âˆ\pounds \mathrm{B})=\frac{P(\mathrm{A}\text{and}\mathrm{B})}{P(\mathrm{B})}=\frac{0.000099700225}{0.000253605625}≈0.3931$

Thus,

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