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. Here are approximate probabilities of positive and negative EIA outcomes when the blood tested does and does not actually contain antibodies to HIV:22 Suppose that $1\%$of a large population carries antibodies to HIV in their blood.

(a) Draw a tree diagram for selecting a person from this population (outcomes: antibodies present or absent) and for testing his or her blood (outcomes: EIA positive or negative).

(b) What is the probability that the EIA test is positive for a randomly chosen person from this population?

(c) What is the probability that a person has the antibody given that the EIA test is positive?

Short answer

Part (b) P (P) = $1.5925\%$

Part (c) P (A|P) = $0.6270$

Part (a) The tree is

Step-by-step solution

Part (a) Step 1. Given Information

The table below depicts the presence of anti-HIV antibodies.

Part (a) Step 2. Explanation

When we know that antibodies are present or absent, the table shows the chances of having a positive or negative test result.

Antibodies are also known to be carried by $1\%$of the population.

Part (b) Step 1. Calculation

Let

$$\begin{gathered} A=Anti-bodiespesent, \\ {A}^C=Anti-bodiesabsent \\ P=Positive, \\ {P}^C=negative \end{gathered}$$

Use the complementary rule to your advantage:

$P(A^C)=1-P\left(A\right)=1-0.01=0.99$

Apply the following generic multiplication rule:

$P(PandA)=P\left(A\right)\times P\left(P\right|A)=0.01\times 0.9985=0.009985$

$P(Pand{A}^C)=P(A^C)\times P(P/A^C)=0.99\times 0.006=0.00594$

Mutually exclusive events should be added together using the addition rule (as the vehicles cannot be filled with two types at the same time).

$$\begin{gathered} P\left(P\right)=P(PandA)+P(Pand{A}^C) \\ =0.009985+0.00549 \\ =0.015925 \\ =1.5925\% \end{gathered}$$

Part (c) Step 1. Concept

Conditional probability: $P\left(A\right|B)=P(AandB)/P(B)$

Part (c) Step 2. Calculation

Let

$$\begin{gathered} A=Anti-bodiespresent, \\ {A}^C=Anti-bodiesabsent \\ P=Positive, \\ {P}^C=negative \end{gathered}$$

P (P) = 1.5925%

conditional probability: $P\left(A\right|B)=P(AandB)/P(B)$

As a result, we have: As a result, we have:

$$\begin{gathered} P\left(A\right|P)=P(AandP)/P(P) \\ =0.009985/0.015925 \\ =0.6270 \end{gathered}$$