Answer:
1. If an individual from Swaziland has tested positive, what is the probability that he carries HIV ?
P=0.8249 or 82.49%
2. If an individual from Swaziland has tested negative, what is the probability that he is HIV free ?
P=0.9988 or 99.88%
Step-by-step explanation:
Make the conditional probability table:
Individual
Infected Not infected
ELISA
Positive
Negative
Totals
The probability of an infected individual with a positive result from the ELISA is obtained from multiplying the probability of being infected (25.9%) with the probability of getting a positive result in the test if is infected (99.7%), the value goes in the first row and column:
P=0.259*0.997=0.2582 or 25.82%
Individual
Infected Not infected Totals
ELISA
Positive 25.82%
Negative
Totals
The probability of a not infected individual with a negative result from the ELISA is obtained from multiplying the probability of not being infected (100%-25.9%=74.1%) with the probability of getting a negative result in the test if isn't infected (92.6%), the value goes in the second row and column:
P=0.741*0.926=0.6862 or 68.62%
Individual
Infected Not infected Totals
ELISA
Positive 25.82%
Negative 68.62%
Totals
In the third row goes the total of the population that is infected (25.9%) and the total of the population free of the HIV (74.1%)
Individual:
Infected Not infected Totals
ELISA
Positive 25.82%
Negative 68.62%
Totals 25.9% 74.1%
Each column must add up to its total, so the probability missing in the first column is 25.9%-25.82%=0.08%, and the ones for the second column is 74.1%-68.62%=5.48%.
Individual
Infected Not infected Totals
ELISA
Positive 25.82% 5.48%
Negative 0.08 68.62%
Totals 25.9% 74.1%
Individual
The third column is filled with the totals of each row:
Infected Not infected Totals
ELISA
Positive 25.82% 5.48% 31.3%
Negative 0.08 68.62% 68.7%
Totals 25.9% 74.1% 100%
The probability A of tested positive is 31.3% and the probability B for tested positive and having the virus is 25.82%, this last has to be divided by the possibility of positive.
P(B/A)=0.2582/0.313=0.8249 or 82.49%
The probability C of tested negative is 68.7% and the probability D for tested negative and not having the virus is 68.62%, this last has to be divided by the possibility of negative.
P(D/C)=0.6862/0.687=0.9988 or 99.88%
