Only 0.1% of the individuals in a certain population have a particular disease (an incidence rate of 0.001). Of those who have the disease, 95% test positive when a certain diagnostic test is applied. Of those who do not have the disease, 90% test negative when the test is applied. Suppose that an individual from this population is randomly selected and given the test: a) Calculate P(positive test) b) Calculate P(has disease | positive test)

The probability will be "0.017699".

Given:

Incident rate,

Now,

→

then,

→

or,

By substituting the above values, we get

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We're given

P(has disease) = 0.001

P(positive test | has disease) = 0.95

P(negative test | does not have disease) = 0.90

(a) By the law of total probability,

P(positive test) = P(positive test AND has disease) + P(positive test AND does not have disease)

Recall that the probability of an event A conditioned on another event B is

P(A | B) = P(A AND B) / P(B)

so that

P(positive test) = P(positive test | has disease) * P(has disease) + P(positive test | does not have disease) * P(does not have disease)

P(positive test) = 0.95 * 0.001 + (1 - 0.90) * (1 - 0.001)

P(positive test) ≈ 0.10

(b) From the definition of conditional probability, we hvae

P(A | B) = P(A AND B) / P(B) = P(B | A) P(A) / P(B)

so that

P(has disease | positive test) = P(positive test | has disease) * P(has disease) / P(positive test)

P(has disease | positive test) = 0.95 * 0.001 / 0.10

P(has disease | positive test) ≈ 0.0094

5/6

step-by-step explanation:

if i remember correctly with perpendicular stuff you just take away the negative

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !