Probability 19
2009-07-10 10:14 pm
A doctor is certain that a patient has the symptoms of disease A but knows that diseases B and C show similar symptoms. In the general population these disease are found in the ratio 5:1:2 . The doctor decides to use a new test and makes 4 independent trials of this test. Find the probability that the patient has disease A if 3 of the trials give positive results and 1 gives a negative one. It is known that the test reacts positively to disease A,B and C with probabilities 0.9 , 0.5 and 0.3 respectively.
回答 (2)
2009-07-11 2:39 am
✔ 最佳答案
Let A, B, C represent the events the patient is of disease A, B, C respectively.Let X be the event that the tests result in 3 positive and 1 negative.
Pr(A|X) = Pr(A∩X)/Pr(X)
Pr(A∩X) = (5/8)(4C3)(0.9)3(0.1) = 0.182
Pr(X) = (5/8)(4C3)(0.9)3(0.1) + (1/8)(4C3)(0.5)3(0.5) + (2/8)(4C3)(0.3)3(0.7) = 0.232
Pr(A|X) = 0.182/0.232 = 0.784
2009-07-11 1:21 am
Pr (A | 3 pos & 1 neg)
= Pr (A & 3 pos & 1 neg) / [Pr (A & 3 pos & 1 neg) + Pr (B & 3 pos & 1 neg) + Pr (C & 3 pos & 1 neg)]
= (0.93 * 0.1) / [(0.93 * 0.1) + (0.53 * 0.5) + (0.33 * 0.7)]
= 0.0729 / (0.0729 + 0.0625 + 0.0189)
= 729 / 1543
= 0.4725
= Pr (A & 3 pos & 1 neg) / [Pr (A & 3 pos & 1 neg) + Pr (B & 3 pos & 1 neg) + Pr (C & 3 pos & 1 neg)]
= (0.93 * 0.1) / [(0.93 * 0.1) + (0.53 * 0.5) + (0.33 * 0.7)]
= 0.0729 / (0.0729 + 0.0625 + 0.0189)
= 729 / 1543
= 0.4725
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