Please help me with a step by step solution to this question.

Question

An X-ray test is used to detect a disease that occurs, initially without any obvious symptoms, in 3% of the population. The test has the following error rates: 7% of people who are disease free have a positive result and 2% of the people who have the disease have a negative result. A large number of people are screened at random using the test, and those with a positive result are examined further. What proportion of people who have the disease are correctly tested?

Answer 1

Note that when you say proportion of people who have the disease are correctly tested, I would assume that you are referring to "how many of the people who actually have the disease were correctly tested." In which case, the answer is 100% - 2% = 98%.

If you mean "how many of the people who were diagnosed as having the disease, who actually do have the disease", then consider the following:

97% of the population are disease free, and 3% have disease. Then, 7% of the 97% disease free people are mis-diagnosed (so their test result is positive). 98% of the 3% who have the disease are correctly diagnosed (test result is positive), so we have the following:

$$.07 * .97 = 0.0679$$ $$.98*.03 = 0.0294$$

You add these up to get the percentage of people who were diagnosed as having the disease. Then you find the fraction of the positively tested people who actually had the disease:

$$\frac{0.0294}{0.0679+0.0294} = \frac{0.0294}{0.0973} = 0.302158273381295 $$

This is approximately 30%. If this is not the answer you were looking for, please clarify the question. Hope this helps!