Probability of malaria

A survey carried out at a certain hospital indicates that the probability that a patient testing positive for malaria is 0.6. What is the probability that two patients were selected at random
(i) one is negative while the other tested positive
(i) both patients tested positive.

Correct answer:

p₁ =  0.48
p₂ =  0.36

Step-by-step explanation:

$$\begin{gathered} p=0.6 \\ {C}_1(2)=\left(\genfrac{}{}{0ex}{}{2}{1}\right)=\frac{2!}{1!(2-1)!}=\frac{2}{1}=2 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{2}{1}\right)\cdot p\cdot (1-p)=2\cdot 0.6\cdot (1-0.6)=0.48 \end{gathered}$$

$p_2=p^2=0.6^2=0.36$

Try another example