Probability questions. Answers are here but I don't know how to do the work so please provide all steps. Thank you!!
1a) P(accident) =P(accident prone)*P(accident|accident prone)+P(not accident prone)*P(accident|not accident prone)=0.3*0.4+(1-0.3)*0.2=0.26
b) P(accident prone|accident)=P(accident prone)*P(accident|accident prone)/P(accident)
=0.3*0.4/0.26=0.46154
2)
P(tested positive )=P(have disease)*P(tested positive|have disease)+P(not have disease)*P(tested positive|not have disease)=0.005*0.95+(1-0.005)*(0.01)=0.0147
therefore P(have disease|tested positive)
=P(have disease)*P(tested positive|have disease)/P(tested positive)
=0.005*0.95/0.0147 =0.323
Probability questions. Answers are here but I don't know how to do the work so please...
Question 10: (10 marks) blood test is 95 percent effective in detecting a certain disease when it is, in fact, also yields a "false positive" result for 10 percent of the healthy persons A laboratory present. However, the test tested. (That is, if a healthy person is tested, then, with probability 0.10, the test result will imply he or she has the disease.) If 0.7 percent of the population actually has the disease, what is the probability a person has...
3) Suppose the probability of having a disease is .001, and a blood test is 100% effective in detecting the disease when the person has the disease. However, the test yields a "false positive" for 1% of healthy persons tested. What is the probability a person has the disease given that his test result is positive? Is the answer approximately a).44 b).01 c).09 d).90
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It’s known that 2 % of people in a certain population have the disease. A blood test gives a positive result (indicating the presence of disease) for 95% of people who have the disease, and it is also positive for 3% of healthy people. One person is tested and the test gives positive result. a. If the test result is positive for the person, then the probability that this person actually has a disease is _________ b. If the test...
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Show your work for each problem. 1) The probability of a driver will have an accident in 1 month equals .02. Find the probability that in 100 months he will have at least one accident. Find the solution exactly and approximately using another distribution. 2) Let X(t) Acoswt 120 where w is a constant and A is a uniform random variable over(0,1). The autocorrelation R, (5,1) is a) b) Coswt - s) cos(ws) cos(wt) 3 دیا 3) Suppose the probability...
A rare but serious disease, D, has been found in 0.01 percent of a certain population. A test has been developed that will be positive, p, for 98 percent of those who have the disease and be positive for 3 percent of those who do not have the disease. Find the probability that a person tested as positive does not have the disease.
I believe I have obtained the correct answers for this, however, I just want to make sure I did the steps correctly. Any help is appreciated, Thank You! oops, sorry! parts b & c (all parts) Suppose a rare, but fatal, disease affects 1 person in n. Researchers have developed a powerful diagnostic test for this disease, but it has a margin of error. Everyone who has the disease will test positive, but B E (0,1) who do not have...
Need details please. Thanks for helping me out ! positive result give that a person 14.A diagnostic test has a probability of 0.90 of giving a suffering from a certain disease, and a probability of 0.15 of give a (false) positive give that the patient is a non-sufferer. It is estimated that 1% of the population are sufferers of this particular disease. Suppose that the test is now administered to a person about whom we have no relevant information relating...
probabilities I know from given problem: .99 have disease AND Test + therefore... .01 have disease AND Test - .02 do not have disease AND Test + therefore... .98 do not have disease AND Test - .10 of TOTAL population HAVE Disease therefore... .90 of TOTAL population DO NOT HAVE Disease. what I thought I would have to do to get what is being asked is P(have disease | tests +) = P(Have disease AND Test +) / P(test +)...