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According to complement rule,

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

According to general multiplication rule,

$P(\mathrm{A}\text{and}\mathrm{B})=P(\mathrm{A}鈭�\mathrm{B})=P(\mathrm{A})脳P(\mathrm{B}鈭�\mathrm{A})=P(\mathrm{B})脳P(\mathrm{A}鈭�\mathrm{B})$

According to addition rule for mutually exclusive event,

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

Conditional probability definition:

$P(\mathrm{B}鈭�\mathrm{A})=\frac{P(\mathrm{A}鈭�\mathrm{B})}{P(\mathrm{A})}=\frac{P(\mathrm{A}\text{and}\mathrm{B})}{P(\mathrm{A})}$

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,

$P(\mathrm{P}鈭�\mathrm{A})=0.9985$

Probability for Antibodies present Negative EIA test,

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

Probability for Antibodies absent Positive EIA test,

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

Probability for Antibodies absent Negative ElA test,

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

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,

$P\left(\mathrm{P}\text{and}\mathrm{A}\right)=P(\mathrm{A})脳P(\mathrm{P}鈭�\mathrm{A})=0.01脳0.9985=0.009985$

Probability for Positive EIA test and Antibodies absent,

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

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:

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

Thus,

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