數統-Probability&distribution一問~

2010-10-26 11:12 pm
以下是HKALAM II 1997年既題目,我剩係知個答案,但唔知點計,希望大家可以教曉我啦~^^ THX

It is known that 4% of the adults in a city have coronary heart disease.
A health centre provides a medical test to diagnose this disease.
Extensive study has shown that on receiving the test,

(1) 95% of adults having coronary heart disease will be correctly diagnosed as
having the disease ; and

(2) 3% of healthy adults will be incorrectly diagnosed as having the disease.

Question:
(a), Find the probability that a randomly selected adult receiving the test is
diagnosed as having coronary heart disease.

(b), Mr Chan is diagnosed by the test as having coronary heart disease.
Find the probability that he actually does not have the disease.

(c), In a certain day, five adults receive the test. Find the probability that only one
of them is incorrectly diagnosed.

P.S ANS for (a),(b) and (c):

(a)= 0.0668
(b)= 0.4311
(c)= 0.1359
更新1:

RE:回答者: 壞人 首先好多謝你的幫忙,不過可惜的是你的解答方式並非為數統科所應用的定律或公式,所以未能解答得到我的問題。不過仍然很多謝你的指導。

回答 (2)

2010-10-27 1:57 am
✔ 最佳答案
(a) P(Having heart disease) = 0.04, so P(No heart disease) = 1 - 0.04 = 0.96.
P(Correctly diagnosed as having disease| have heart disease) = 0.95,
P(Incorrectly diagnosed as having disease| not having disease) = 0.03 [Note : Healthy adult means not having disease]
By Law of Total Probability :
P(Diagnosed as having heart disease) = P(Diagnosed as having heart disease|have heart disease)P(Having heart disease) + P(Diagnosed as having heart disease|not having heart disease)P(Not having heart disease)
= 0.95 x 0.04 + 0.03 x 0.96 = 0.038 + 0.0288 = 0.0668.
(b) By Bayes' Theorem,
P(No heart disease|Diagnosed as having heart disease)
= P(Diagnosed as having heart disease|No heart disease)P(No heart disease)/P(Diagnosed as having heart disease)
= 0.03 x 0.96/0.0668 = 0.4311.
(c) By Law of Total Probability,
P(Incorrectly diagnosed as having heart disease) = P(Diagnosed as having heart disease|No heart disease)P(No heart disease) + P(Diagnosed as not having heart disease|Have heart disease) P(Have heart disease)
= 0.03 x 0.96 + ( 1 - 0.95) x 0.04 = 0.0288 + 0.002 = 0.0308.
Let X = No. of adults incorrectly diagnosed out of 5 adults.
This fulfill the Binomial distribution with n = 5 and p = 0.0308
so X ~ B(5, 0.0308)
P(Only 1 is incorrectly diagnosed) = P(X = 1) = (5C1)(0.0308)(1 - 0.0308)^4
= 5 x 0.0308 x 0.9692^4 = 0.1359
2010-10-26 11:48 pm
(a) Suppose x be the number of adults that have heart disease and get correct result from the test.
Then adults that have heart disease but get incorrect result from the test
= x(0.05/0.95)
= x/19
Adults that do not have heart disease but get incorrect result from the test
= (x + x/19)(0.96/0.04)(0.03)
= 72x/95
Adults that do not have heart disease and get correct result from the test
= (x + x/19)(0.96/0.04)(0.97)
= 2328x/95
Required probability
= (x + 72x/95)/(x + x/19 + 72x/95 + 2328x/95)
= 0.0668

(b) Required probability
= (72x/95)/(72x/95 + x)
= 0.4311

(c) Probability that an adult get incorrect result
= (x/19 + 72x/95)/(x + x/19 + 72x/95 + 2328x/95)
= 0.0308
Required probability
= 5(0.0308)(1-0.0308)^4
= 0.1359
參考: me


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