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andrezito [222]
1 year ago
13

It is estimated that approximately 8.23% Americans are afflicted with diabetes. Suppose that a certain diagnostic evaluation for

diabetes will correctly diagnose 98% of all adults over with diabetes as having the disease and incorrectly diagnoses 3.5% of all adults over without diabetes as having the disease.
a) Find the probability that a randomly selected adult over does not have diabetes, and is diagnosed as having diabetes (such diagnoses are called "false positives").
b) Find the probability that a randomly selected adult of is diagnosed as not having diabetes.
c) Find the probability that a randomly selected adult over actually has diabetes, given that he/she is diagnosed as not having diabetes
Mathematics
1 answer:
allsm [11]1 year ago
7 0

The probabilities in this problem are given as follows:

a) False positive: 0.0321 = 3.21%.

b) Diagnosed as not having diabetes: 0.8872 = 88.72%.

c) Actually has diabetes, if diagnosed as not having: 0.0019 = 0.19%.

<h3>What is Conditional Probability?</h3>

Conditional probability is the probability of one event happening, considering a previous event. The formula is given as follows:

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

In which the parameters are described as follows:

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

For item a, we have that:

  • 100 - 8.23 = 91.77% of the people do not have diabetes.
  • Of those, 3.5% are diagnosed with diabetes.

Hence the probability of a false positive is given as follows:

p = 0.9177 x 0.035 = 0.0321 = 3.21%.

For item b, the percentage of people who is not diagnosed as having diabetes is divided as:

  • 96.5% of 91.77% (do not have diabetes).
  • 2% of 8.23% (have diabetes).

Hence the probability is:

P(A) = 0.965 x 0.9177 + 0.02 x 0.0823 = 0.8872 = 88.72%.

For item c, we find the conditional probability, as follows:

P(A \cap B) = 0.02 \times 0.0823 = 0.001646

Then:

P(B|A) = 0.001646/0.8872 = 0.0019 = 0.19%.

More can be learned about probabilities at brainly.com/question/14398287

#SPJ1

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