Question

1. a disease is present in 8% of the population. a test will return a positive result 90% of the time if the disease is present and will return a negative result 95% of the time if the disease is not present.

(a) draw a tree - diagram for this process.
(b) what is the probability that a randomly selected person has the disease given that the result of the test is positive?
(c) what is the probability that a randomly selected person has the disease given that the result of the test is negative?

Explanation:

Step1: Define events

Let $D$ be the event that a person has the disease, and $\overline{D}$ be the event that a person does not have the disease. Let $+$ be the event that the test is positive and $-$ be the event that the test is negative. We know $P(D)=0.08$, $P(\overline{D}) = 1 - P(D)=0.92$, $P(+|D)=0.9$, $P(-|D)=1 - P(+|D)=0.1$, $P(-|\overline{D}) = 0.95$ and $P(+|\overline{D})=1 - P(-|\overline{D})=0.05$.

Step2: Draw the tree - diagram (description)

The first - level branches are for the presence or absence of the disease: one branch for $D$ with probability $P(D) = 0.08$ and another for $\overline{D}$ with probability $P(\overline{D})=0.92$. From the $D$ branch, there are two secondary branches: one for $+$ with probability $P(+|D)=0.9$ and one for $-$ with probability $P(-|D)=0.1$. From the $\overline{D}$ branch, there are two secondary branches: one for $+$ with probability $P(+|\overline{D}) = 0.05$ and one for $-$ with probability $P(-|\overline{D})=0.95$.

Step3: Calculate $P(+)$ using the law of total probability

By the law of total probability, $P(+) = P(+|D)P(D)+P(+|\overline{D})P(\overline{D})$.
Substitute the known values: $P(+)=(0.9\times0.08)+(0.05\times0.92)=0.072 + 0.046=0.118$.

Step4: Calculate $P(D|+)$ using Bayes' theorem

By Bayes' theorem, $P(D|+)=\frac{P(+|D)P(D)}{P(+)}$.
Substitute the values: $P(D|+)=\frac{0.9\times0.08}{0.118}=\frac{0.072}{0.118}\approx0.61$.

Step5: Calculate $P(-)$ using the law of total probability

$P(-)=P(-|D)P(D)+P(-|\overline{D})P(\overline{D})=(0.1\times0.08)+(0.95\times0.92)=0.008 + 0.874 = 0.882$.

Step6: Calculate $P(D| -)$ using Bayes' theorem

By Bayes' theorem, $P(D| -)=\frac{P(-|D)P(D)}{P(-)}$.
Substitute the values: $P(D| -)=\frac{0.1\times0.08}{0.882}=\frac{0.008}{0.882}\approx0.009$.

Answer:

(a) The tree - diagram has first - level branches for disease presence ($D$ with $P(D) = 0.08$) and absence ($\overline{D}$ with $P(\overline{D}) = 0.92$). From $D$, there are branches for positive ($+$ with $P(+|D)=0.9$) and negative ($-$ with $P(-|D)=0.1$) test results. From $\overline{D}$, there are branches for positive ($+$ with $P(+|\overline{D}) = 0.05$) and negative ($-$ with $P(-|\overline{D})=0.95$) test results.
(b) $P(D|+)\approx0.61$
(c) $P(D| -)\approx0.009$