Does somebody knows how to do this problem?
1 answer:
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Complete Question
Suppose there was a cancer diagnostic test was 95% accurate both on those that do and 90% on those do not have the disease. If 0.4% of the population have cancer, compute the probability that a particular individual has cancer, given that the test indicates he or she has cancer.
Answer:
The probability is
Step-by-step explanation:
From the question we are told that
The probability that the test was accurate given that the person has cancer is
The probability that the test was accurate given that the person do not have cancer is
The probability that a person has cancer is
Generally the probability that a person do not have cancer is
=>
=>
Generally the probability that a particular individual has cancer, given that the test indicates he or she has cancer is according to Bayes's theorem evaluated as
=>
=>
Answer:
0.912
Step-by-step explanation:
<em>The complete question is attached.</em>
<em />
"OR" in probability means "ADDITION". So we find the individual probabilities and "ADD" them.
P(Plan A) would be the total people for plan A (add up column of Plan A) divided by total number of people [add up all the numbers].
Looking at the table, we have:
P (Plan A) = 38/68
Now,
P(40-49 Group) can be found by adding up the row of 40-49 age group and dividing by the total. So, we have:
P(40 - 49 Group) = 24/68
Now, adding up both:
38/68 + 24/68 = 62/68 = 0.912
