A pregnant woman lives in an area where the Zika virus is fairly rare: $1$ in $1000$ people have it. Still, she is concerned; so she gets tested. There is a good but not perfect test for the virus - it gives a positive reading with probability $0.99$ if the person has the virus and a positive reading with probability $0.05$ if the person does not. Her reading is positive. What is the probability that she has the virus?
(a) About $2\%$
(b) About $10\%$
(c) About $ 80\%$
(d) About $20\%$
On
I would recommend you to use a fourfold table. They're extremly helpful when it comes to dealing with probability. Suppose we are working with $100000$ people. Let $Z=$"has Zike", $\overline Z$="doesn't have Zike", $T$="positive test", $\overline T$="negative test". Then
\begin{array}{c|lcr} & Z & \overline Z & \sum\\ \hline T & \color{red}{99} & 4995 & 5094 \\ \overline T & 1 & 94905 & 94906 \\ \sum& 100 & 99900 & 100000 \end{array}
The pregnant woman's test states that she has Zika. So we want to calculate the probability $P$, that among the $5094$ people who received a positive test, she belongs to the $99$ who indeed have Zika. Thus
$$P=\frac{99}{5094}\approx 2\%$$
Use conditional probability!
First $1$ in $1000$ is the same as $0.1\%$, then calculate the probability that she has the disease:
$$P(Diseased)=(0.001*0.99)+(0.999*0.05)=0.05094$$
Next we use conditional probability:
$$P(Diseased|Positive)=\frac{0.001*0.999}{0.05094}$$
So this is equal to $0.0196.....$ which is $1.96\%$. Therefore a) is the correct answer!