The prevalence of a certain disease is 0.09. A diagnostic test has been developed to test for the disease in question which returns a positive reading with probability 0.9 amongst those where it is known they have the disease and returns a negative result with probability 0.96 for those that are known not to have the disease. A person is suspected to have the disease, takes the test and the test comes back positive. What is the probability that they have the disease given that the test came back positive ?

The probability that they have the disease given that the test came back positive is 0.6899.

Step-by-step explanation:

Let A represent the event that the prevalence of a certain disease is 0.09

Let D be the event in which those patients are tested who already have the disease then the probability of B

P (D) = 0.9

Let F be the event in which those patients are tested who do not have the disease then the probability of F

P (F) = 0.96

The tree diagram helps explain the events and their probabilities.

A 0.09 0.9 (D)= P(D/A) (positive test)

       ⇅0.1 ( negative test)

       ⇅

1 ⇅

       ⇅

       ⇅

B 0.91 0.96 (F)= P(F/ B) (negative test)

0.04 P (G/B) (positive test)

By Bayes Theorem

P (A/ D)= P (A). P(D/A)/ P (A). P(D/A)+P (B) P(G/ B)

P (A/ D)= 0.09 (0.9)/ 0.09 (0.9)+ (0.91)(0.04)

P (A/ D)= 0.081/ 0.081+0.0364

P (A/ D)= 0.081/ 0.1174

P (A/ D)= 0.6899

The probability that they have the disease given that the test came back positive is 0.6899.

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.

8496 cars

step-by-step explanation: