Question

A firm has developed a new test for COVID-19. If someone has COVID-19, the test finds it 93% of the time. If someone does not have COVID-19 the test returns a positive COVID-19 8% of the time. If the 30% of the populations has COVID-19, what is the probability of someone selected at random having COVID-19 if they tested positive? Give your answer to four decimal places.

Please help and show work.

Answer #1

Lets define C as the event which denotes someone having covid-19 then C' denotes the event in which someone doesn't have covid-19. Also, define +ve as the event where test results in a positive and -ve if the test results negative.

According to the information given then we can construct the tree diagram, since

P(C)= 30/100=0.3 and P(C')= 1-0.3=0.7,

P(+ve | C)= 93/100= 0.93, so P(-ve | C)= 0.07

P(+ve | C')= 8/100= 0.08, and P(-ve | C')= 0.92 as represented in the following tree diagram where the extended branches show the conditional probabilities.

Now, P( +ve )= P(+ve | C)P(C) + P(+ve | C')P(C') = 0.3*0.93+ 0.7*0.08 = 0.335

P(+ve and C)= P(+ve| C)P(C)= 0.279

correct upto four decimal places.

So the probability of someone selected at random having COVID-19 if they tested positive = 0.8328

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