Answer :
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
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
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
- P(B|A) is the probability of event B happening, given that A happened.
- [tex]P(A \cap B)[/tex] 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:
[tex]P(A) = 0.95(0.08) + 0.02(0.92) = 0.0944[/tex]
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, [tex]P(A) = 0.0944[/tex].
The probability of both a positive test and having the disease is:
[tex]P(A \cap B) = 0.95(0.08)[/tex]
Hence, the conditional probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.95(0.08)}{0.0944} = 0.8051[/tex]
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 https://brainly.com/question/14398287