Solve Engineering Statistics Homework: Positive result on 2 Tests

Click For Summary
SUMMARY

The probability of having a disease given a positive result on two successive tests is calculated to be 0.9235, based on the initial probability of the disease being 0.005 and the test's sensitivity of 0.98 and false positive rate of 0.02. The initial probability of having the disease given one positive test result is 0.1976. The solution involves applying Bayes' theorem and considering the independence of test results, rather than simply multiplying probabilities. The correct approach requires a detailed understanding of conditional probabilities and the law of total probability.

PREREQUISITES
  • Understanding of Bayes' theorem
  • Knowledge of conditional probability
  • Familiarity with statistical independence
  • Ability to calculate probabilities from test sensitivity and specificity
NEXT STEPS
  • Study Bayes' theorem applications in medical testing
  • Learn about conditional probability calculations
  • Research the law of total probability
  • Practice problems involving statistical independence in probability
USEFUL FOR

Students in statistics, healthcare professionals involved in diagnostic testing, and anyone interested in understanding probability in medical contexts.

wolfmanzak
Messages
25
Reaction score
0

Homework Statement



People in a given community who have a certain disease is 0.005. A test is available to diagnose the disease. If a person has the disease, the probability that the test will produce a positive signal is 0.98. If a person does not have the disease, the probability that the test will produce a positive signal is 0.02.

If a person tests positive on two successive tests, what is the probability that they have the disease?

Homework Equations



I was assuming statistical independence, though the problem doesn't necessarily state that, so I was contemplating just multiplying the two probabilities.


The Attempt at a Solution



I know from a previous part of the problem that the probability of having the disease, given a positive test result is .1976. I was going to multiply .1976*.1976 to get the answer for this question but according to the solution my TA provided that's wrong. His solution is .9235, which when I think about it, makes sense. But I don't know what he did to get to that value?
 
Physics news on Phys.org
wolfmanzak said:
I know from a previous part of the problem that the probability of having the disease, given a positive test result is .1976. I was going to multiply .1976*.1976 to get the answer for this question but according to the solution my TA provided that's wrong. His solution is .9235, which when I think about it, makes sense. But I don't know what he did to get to that value?
Are you able to show the working to obtain .1976 as the answer to the earlier part? If you can do part (a) then that same method, extended to two tests, will give the answer to part (b).

http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon2.gif So, work at showing how to find .1976 first.
 
Last edited by a moderator:

Similar threads

  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K