0

Suppose that 1 in 3 people showing symptoms of dry cough, fatigue (tiredness)
and fever have COVID19.There is a test for COVID-19 that gives a positive result 98.5%
of the time when given to someone with the disease. When given to someone without the disease.
80% of the time it gives a negative result.

  1. What is the probability that a randomly selected person showing symptoms of dry cough, fatigue (tiredness) and fever has COVID-19 if the test returns a positive result?
  2. What is the probability that a randomly selected person showing symptoms of dry cough, fatigue (tiredness) and fever does not have COVID-19 if the test returns a negative result?

I approached the problem following way but got stuck
Let +ve denote the people who test positive.
And covid denote people who actually have covid.
Now from the given statements we can say that
P(+ve | covid) = .985 -- Means Given a person has
covid the probability that the test result will be +ve is .985
P(-ve | non-covid) = .8
P(covid | symptoms) = 1/3

Now the Q1. asks to find out P(+ve | symptoms)
P(+ve | symptoms ) = P(symptoms | +ve) x P(+ve) / P(symptoms)

Now the test gives +ve 98.5% of time for one with covid and
when given to somone without covid it gives +ve 20% of the time.
P(+ve | non-covid) =.2 Dont know how to proceed further

ryang
  • 38,879
  • 14
  • 81
  • 179
nisha aunty
  • 11

1 Answers1

1

I will answer the first questions, the second is very similiar.

$$p(\text{covid}|\text{positive}) = \frac{p(\text{positive}|\text{covid})p(\text{covid})}{p(\text{positive}|\text{covid})p(\text{covid})+ p(\text{positive}|\text{not covid})p(\text{not covid})} =\\ \frac{0.985\times0.33}{0.985\times0.33 + 0.2\times0.66} \sim 70\% $$

This is why doctors make patients take more tests if a test is known to give false positives. See pic for a geometric interpretation for this kind of problems: think about plugging a number of random event at top.

Bye

deppep
  • 78
  • in the above you have assumed that p(covid) = 0.33 . Although the question mention its the probabilty of people showing symptoms + covid. So there could be people who have covid and not have symptoms. so p(covid) = p(covid with symptoms) + P(covid without symptoms). so what I understand is** p(covid with symptoms) = .33**. Correct me – nisha aunty Dec 19 '20 at 13:22
  • 1
    yes of course. for brevity sake i write $p(\text{covid})$ and $p(\text{not covid})$ basically meaning $p(\text{symptomatic person has covid})$ and $p(\text{symptomatic person does not have covid})$. we have no informations about asymptomatic covid here @nishaaunty – deppep Dec 19 '20 at 13:26