
What is the probability of the ‘person to be tested as COVID positive’ given that ‘he is actually not having COVID’?
Answer & explanation
Correct answer: option 1
Let,
E: The event that person selected has COVID.
F: The event that person selected does not have COVID.
G: The event that person is tested positive.
$P\left(\frac{\text{tested covid}}{\text{has covid}}\right)=P(\frac{G}{E})=90\%$
$=0.9$
$P\left(\frac{\text{tested covid positive}}{\text{does not have covid}}\right)=1\%=0.01$
$\text{person not having covid}=1-P(person covid)$
$P(F)=1-P(E)$
$=1-0.001=0.999$
$P\left(\frac{\text{having covid}}{\text{covid positive}}\right)=P(\frac{E}{G})$
$=\frac{P(E).P(G|E)}{P(E)P(G|E)+P(F)P(G|F)}$
$=\frac{0.001×0.9}{0.001×0.9+0.999×0.01}$
$=0.083$
$\text{P(person selected will be diagnosed as covid positive)} = P(F)×P(G|F)+P(E)P(G|E)$
$=0.001×0.9+0.999×0.01$
$=0.01089$