The probability of A is 0.45. The conditional probability that A occurs given th

Question

The probability of A is 0.45. The conditional probability that A occurs given that B occurs is 0.55. The conditional probability that A occurs given that B occur is 0.65 a. What is the probability that B occurs? b. What is the conditional probability B does not occur given that A does not occurs? It is believed that 3% of a clinic's patients have TB. A particular blood test yields a positive result for 99% of patients with TB, but also shows positive for 3% of patients who do not have TB. One patient is chosen at random from the clinic's patient list and is listed. What is the probability that if the test result is positive, the person actually has TB? Let the random variables X be given by the following probability distribution: P(X = -2) = 1/10, P(X = 0) = 1/10, P(X = 5) = 5/10, P(X = 19) = 1.5/10, P(X = 17) = 1.5/10. a. Find mu and sigma of chi. b. Find P(X greaterthanorequalto 5), and P(mu - sigma

Explanation / Answer

Q1 seems incorrect hence answering Q2.

Answer 2.

P( TB ) = .03 // probability of a person coming to clinic having TB

P(TEST/TB ) = .99 // probability of test being positive for a TB patient.

P(TEST / !TB ) = .03 // probability of test being positive for a non TB patient.

We need to find P( TB / TEST) .

P( TB / TEST ) = P(TEST / TB ) * P(TB) / P(TEST) -- (1)

P(TEST / !TB) = P( !TB / TEST ) * P(TEST) / P( ! TB)   

this gives P( TEST ) = P(TEST/ !TB ) * P( !TB) / P( !TB/TEST) = .03* .97 / P( !TB/TEST)

also, P(TB/TEST) + P( !TB / TEST ) = 1

so, P(TEST) = .03 * .97 / ( 1 - P( TB/TEST))

putting this in (1)

P(TB/TEST) = .99 * .03 * (1 - P(TB/TEST)) / ( .03 * .97) = 99(1-P(TB/TEST)) / 97

rearranging above gives us : 196 P(TB/TEST) = 99

PB( TB/TEST ) = 99/196

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