Mathematics, 16.10.2020 04:01, ricardoamora54

In a certain community, 8% of all adults over 50 have diabetes. If a health service in this community correctly diagnoses 95% of all persons with diabetes as having the disease and incorrectly diagnoses 2% of all persons without diabetes as having the disease, find the probabilities that a) the community health service will diagnose an adult over 50 as having diabetes.
b) a person over 50 diagnosed by the health service as having diabetes actually has the disease.

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answered: bahjaMcclatchey1110

Using conditional probability, it is found that there is a:

a) 0.0944 = 9.44% probability that the community health service will diagnose an adult over 50 as having diabetes.

b) 0.8051 = 80.51% probability that a person over 50 diagnosed by the health service as having diabetes actually has the disease.

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened. $P(A \cap B)$ is the probability of both A and B happening. P(A) is the probability of A happening.

Item a:

The percentages of a positive test is:

95% of 8%(have diabetes).2% of 92%(do not have diabetes).

Hence:

P(A) = 0.95(0.08) + 0.02(0.92) = 0.0944

0.0944 = 9.44% probability that the community health service will diagnose an adult over 50 as having diabetes.

Item b:

Event A: Positive test.Event B: Has the disease.

From item a, P(A) = 0.0944 .

The probability of both a positive test and having the disease is:

$P(A \cap B) = 0.95(0.08)$

Hence, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.95(0.08)}{0.0944} = 0.8051$

0.8051 = 80.51% probability that a person over 50 diagnosed by the health service as having diabetes actually has the disease.

A similar problem is given at link

answered: dontcareanyonemo

Answer and Step-by-step explanation:

Solution:

Given:

Let A is the event of a person over 50 diagnosed by health service.

B1 is the event that an adult over 50 actually has diabetes.

B2 is the event that an adult over 50 does not have diabetes.

P (B1) = 8% = 0.08

And P (B2) = 1 – P (B1)

= 1 – 0.08

= 0.92

Correctly diagnose of all persons with diabetes= 95%

P (A/B1) = 0.95

Incorrectly diagnose of all persons without diabetes= 2%

P (A/B2) = 0.02

(a)) the community health service will diagnose an adult over 50 as having diabetes.

P (B1) P (A/B1) = (0.08) (0.95) = 0.076

P (B2) p (A/B2) = (0.92) (0.02) =0.0184

P (B1) P (A/B1) + P (B2) p (A/B2) = 0.076 + 0.0184

= 0.0944

(b) A person over 50 diagnosed by the health service as having diabetes actually has the disease.

By using formula:

P (B1/A) = P (B1) P (A/B1) / P (B1) P(A/B1) + P (B2) P (A/B2)

Put all the given values:

= (0.08) (0.95) / (0.08) (0.95) + (0.92) (0.02)

=0.076 / 0.076 + 0.0184

=0.076 / 0.0944

= 0.8050

answered: Guest

Ithink it’s b, hope this .

answered: Guest

the answer is simply just 6n add the two together as they have an n

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