The test to detect the presence of respiratory syncytial virus is 97% accurate for a person who has the virus and 99% accurate for a person who does not have the virus. In a given population, 0.55% of the people are infected. The probability that a randomly chosen person gets an incorrect result is
.

The probability that a randomly selected person gets incorrect result is 2.2 × 10⁻⁴

Step-by-step explanation:

The parameters given are;

The accuracy of the test for a person who has the respiratory synctial virus = 97%

The accuracy of the test for a person who does not have the respiratory synctial virus = 99%

We have;

a = TP =

b = FP

c = FN

d = TN

a/(a + c) = 0.97

d/(d + b) = 0.99

a/(a + b) = 0.97*0.0055/(0.97*0.0055 + (1 - 0.99)*(1-0.0055))

PPV = 0.349 = 34.9%

Therefore, we have;

a/(a + c) = 0.97 and

a/(a + b)  =  0.349

0.97(a + c) =0.349(a + b)

(0.97 - 0.349)a = 0.349·b - 0.97·c

a = (0.349·b - 0.97·c)0.621

b × (1 - 0.0055) = (1 - 0.97)×(1 - 0.0055)

b = 1 - 0.97 = 0.03

Similarly,

c = 1 - 0.99 = 0.01

The proportion of the population that have false positive and false negative = 0.03 + 0.01 = 0.04 = 4%

The probability that a randomly selected person gets incorrect result = 0.04×0.0055 = 0.00022.

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!