Chapter 7.3, Problem 7.70E

(a)

To determine

To construct: a tree diagram that models this situation

Explanation of Solution

Given:

Assuming that

  $\begin{array}{l}\text{A=Disease}\\ {\text{A}}^{\text{c}}\text{=No}\text{Disease}\\ \text{B= Positive test}\\ {\text{B}}^{\text{c}}\text{=Negative}\text{test}\end{array}$

  $\begin{array}{l}\text{P}\left(\text{A}\right)\text{=0}\text{.001}\\ \text{P}\left({\text{B}}^{\text{c}}\text{|A}\right)\text{=0}\text{.001}\\ \text{P}\left({\text{B|A}}^{\text{c}}\right)\text{=0}\text{.002}\end{array}$

Graph:

Tree diagram:

(b)

To determine

To Calculate: the probability that a randomly selected person from this area would test positive.

Answer to Problem 7.70E

0.002997

Explanation of Solution

Formula used:

Compliment Rule

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

Multiplication Rule

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

Calculation:

  $\begin{array}{c}\text{P}\left({\text{A}}^c\right)=1-\text{P}\left(\text{A}\right)\\ =1-0.001\\ =0.999\\ P\left(B|A\right)=1-P\left(B^c|A\right)\\ =1-0.001\\ =0.999\end{array}$

  $\begin{array}{c}\text{P}\left(B\cap A\right)=\text{P}\left(\text{A}\right)\times \text{P}\left(\text{B|A}\right)\\ =0.001\times 0.999\\ =0.000999\\ \text{P}\left(B\cap A^c\right)=\text{P}\left({\text{A}}^c\right)\times \text{P}\left({\text{B|A}}^c\right)\\ =0.999\times 0.002\\ =0.001998\end{array}$

  $\begin{array}{c}P\left(B\right)=\text{P}\left(B\cap A\right)+\text{P}\left(B\cap A^c\right)\\ =0.000999+0.001998\\ =0.002997\end{array}$

(c)

To determine

To Calculate: the probability that the person actually has HIV.

Answer to Problem 7.70E

0.3333

Explanation of Solution

Formula used:

Conditional Probability

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

Calculation:

  $\begin{array}{c}P\left(A|B\right)=\frac{\text{P}\left(B\cap A\right)}{P\left(B\right)}\\ =\frac{0.000999}{0.002997}\\ =0.3333\end{array}$