anonymous
  • anonymous
Suppose that for the general population, 1 in 5000 people carries the human immunodeficiency virus (HIV). A test for the presence of HIV yields either a positive () or negative () response. Suppose the test gives the correct answer 99% of the time. What is P[-|H], the conditional probability that a person tests negative given that the person does have the HIV virus? What is P[H|+], the conditional probability that a randomly chosen person has the HIV virus given that the person tests positive?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
P[-|H]= P[- * H] / P[H] .01/(1/5000)
anonymous
  • anonymous
is it right?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
this is a baye's formula problem, right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
easiest way to see the answer is to do it with actual numbers. then we can use the formula, although i have to go soon.
anonymous
  • anonymous
suppose 1,000,000 people are tested. of those 1,000,000, exactly 200 will have the disease (assuming the probabilities are exact) then of those 200, 99% will test positive, that is, 198 will test positive, and 2 will test negative. out of the remaining 999,800 1% or 99,980 will also test positive
anonymous
  • anonymous
that was wrong, sorry. last line should be out of the remaining 999,800 1% or 9,998 will test positive
anonymous
  • anonymous
now you can just go ahead and compute
anonymous
  • anonymous
so that 1% accuracy is shared among those wrongly tested positive and wrongly tested negative or it is each 1%?
anonymous
  • anonymous
1% inaccuracy
anonymous
  • anonymous
that is what i am assuming from the wording, yes
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
usually false positive and false negative are different percents, but in this case you were just given one number
anonymous
  • anonymous
9998+200 test positive, but of those with positive tests, only 198 actually have the disease, so the probability that you have the disease given that the test is positive is only \(\frac{198}{9998+200}\) now if you redo the problem using only decimals instead of 1,000,000 you will get bays formula

Looking for something else?

Not the answer you are looking for? Search for more explanations.