Chapter 7.3, Problem 7.76E

(a)

To determine

To Explain: the false positive and false negative rates for the test.

Answer to Problem 7.76E

  $\begin{array}{c}\text{False negative rate }=0.04\\ \text{False positive rate }=0.0667\end{array}$

Explanation of Solution

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Calculation:

240 of the 250 patients with the disease having a positive test result and therefore only 10 of the 250 patients with the disease had a false negative.

  $\begin{array}{c}\text{False negative rate =}\frac{10}{250}\\ =0.04\end{array}$

50 of the 750 patients without the disease have a positive test result and therefore have a false positive.

  $\begin{array}{c}\text{False positive rate =}\frac{50}{750}\\ =0.0667\end{array}$

(b)

To determine

To Calculate: the probability that the patient actually had the medical condition.

Answer to Problem 7.76E

  $\begin{array}{c}P\left(B\right)=0.29\\ P\left(A|B\right)=0.8276\end{array}$

Explanation of Solution

Given:

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

  $\begin{array}{l}\text{P}\left({\text{B}}^{\text{c}}\text{|A}\right)\text{=0}\text{.04}\\ \text{P}\left({\text{B|A}}^{\text{c}}\right)\text{=}\frac{1}{15}\end{array}$

Formula used:

Compliment Rule

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

Multiplication Rule

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

Conditional Probability

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

Calculation:

250 disease patients out of 1000

  $\begin{array}{c}P\left(A\right)=\frac{250}{1000}\\ =0.25\end{array}$

  $\begin{array}{c}\text{P}\left({\text{A}}^c\right)=1-\text{P}\left(\text{A}\right)\\ =1-0.25\\ =0.75\\ P\left(B|A\right)=1-P\left(B^c|A\right)\\ =1-0.04\\ =0.96\end{array}$

  $\begin{array}{c}\text{P}\left(B\cap A\right)=\text{P}\left(\text{A}\right)\times \text{P}\left(\text{B|A}\right)\\ =0.25\times 0.96\\ =0.24\\ \text{P}\left(B\cap A^c\right)=\text{P}\left({\text{A}}^c\right)\times \text{P}\left({\text{B|A}}^c\right)\\ =0.75\times \frac{1}{15}\\ =0.05\end{array}$

  $\begin{array}{c}P\left(B\right)=\text{P}\left(B\cap A\right)+\text{P}\left(B\cap A^c\right)\\ =0.24+0.05\\ =0.29\end{array}$

  $\begin{array}{c}P\left(A|B\right)=\frac{\text{P}\left(B\cap A\right)}{P\left(B\right)}\\ =\frac{0.24}{0.29}\\ =0.8276\end{array}$