Bayes’ theorem describes the relationships that exist within an array of simple and conditional probabilities. For example: Suppose there is a certain disease randomly found in one-half of one percent (.005) of the general population. A certain clinical blood test is 99 percent (.99) effective in detecting the presence of this disease; that is, it will yield an accurate positive result in 99 percent of the cases where the disease is actually present. But it also yields false-positive results in 5 percent (.05) of the cases where the disease is not present.
The following table shows (in red) the probabilities that are stipulated in the example and (in blue) the probabilities that can be inferred from the stipulated information: P(A) = .005the probability that the disease will be present in any particular person P(~A) = 1—.005 = .995the probability that the disease will not be present in any particular person P(B|A) = .99the probability that the test will yield a positive result [B] if the disease is present [A] P(~B|A) = 1—.99 = .01the probability that the test will yield a negative result [~B] if the disease is present [A] P(B|~A) = .05the probability that the test will yield a positive result [B] if the disease is not present [~A] P(~B|~A) = 1—.05 = .95the probability that the test will yield a negative result [~B] if the disease is not present [~A]
The Essay on Addisons Disease
The disease I’ve been reading and researching about is called Addison’s Disease. Addison’s Disease is an uncommon condition in which the adrenal glands, above the kidneys, don’t make enough of a hormone called cortisol, which helps regulate the body’s use of a protein, carbohydrates and fat, helps maintain blood pressure and cardiovascular function and controls ...
Given this information, Bayes’ theorem allows for the derivation of the two simple probabilities P(B) = [P(B|A) x P(A)] + [P(B|~A) x P(~A)]
= [.99 x .005]+[.05 x .995] = .0547the probability of a positive test result [B], irrespective of whether the disease is present [A] or not present [~A] P(~B) = [P(~B|A) x P(A)] + [P(~B|~A) x P(~A)]
= [.01 x .005]+[.95 x .995] = .9453the probability of a negative test result [~B], irrespective of whether the disease is present [A] or not present [~A] which in turn allows for the calculation of the four remaining conditional probabilities P(A|B) = [P(B|A) x P(A)] / P(B)
= [.99 x .005] / .0547 = .0905the probability that the disease is present [A] if the test result is positive [B] (i.e., the probability that a positive test result will be a true positive) P(~A|B) = [P(B|~A) x P(~A)] / P(B)
= [.05 x .995] / .0547 = .9095the probability that the disease is not present [~A]if the test result is positive [B] (i.e., the probability that a positive test result will be a false positive) P(~A|~B) = [P(~B|~A) x P(~A)] / P(~B)
= [.95 x .995] / .9453 = .99995the probability that the disease is absent [~A] ifthe test result is negative [~B] (i.e., the probability that a negative test result will be a true negative) P(A|~B) = [P(~B|A) x P(A)] / P(~B)
= [.01 x .005] / .9453 = .00005the probability that the disease is present [A] if the test result is negative [~B] (i.e., the probability that a negative test result will be a false negative)
To perform calculations using Bayes’ theorem, enter the probability for one or the other of the items in each of the following pairs (the remaining item in each pair will be calculated automatically).
A probability value can be entered as either a decimal fraction such as .25 or a common fraction such as 1/4. Whenever possible, it is better to enter the common fraction rather than a rounded decimal fraction: 1/3 rather than .3333; 1/6 rather than .1667; and so forth. P(A) or P(~A)
The Essay on Jung Typology Test Result
After taking this Test, I realized that I totally was the same as description of ESTJs. Due to findings from it, I feel more believe in my career choice, and I can also prove my strengths as well as remedy my weaknesses. There are numerous opinions about ESTJs that I think they are exactly. First of all, embracing the values of honesty, dedication and dignity, people with the ESTJ personality type ...
P(B|~A) or P(~B|~A)
P(B|A) or P(~B|A) After the three probability values (one from each pair) have been entered, click the cursor anywhere in the gray field of the table to complete the intermediate calculations, then click the “Calculate” button. ~~Note that no probability value can be less than 0.0 or greater than 1.0.
