At the local bank 60% of the employees have 2 or more children. There
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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:
In which the parameters are described as follows:
- P(B|A) is the probability of event B happening, given that event A happened.
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:
Then:
P(B|A) = 0.001646/0.8872 = 0.0019 = 0.19%.
More can be learned about probabilities at brainly.com/question/14398287
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