Probability question

Question

When doing blood testing for a viral infection, the procedure can be made more efficient and less expensive by combining partial samples of different blood specimens. If samples from four people are combined and the mixture tests negative, we know that all four individual samples are negative. Find the probability of a positive result for four samples combined into one mixture, assuming the probability of an individual blood sample testing positive for the virus is 0.12.

*My tutor didn't know how to solve this problem, we tried to find the probabilities for all the samples "PPNP,PPPP..." etc and that answer didn't work. Just the answer for PPPP didn't work either. Not sure what this question is asking.

Answer 1

samples from 4 samples are combined and test negative

P(+) = .12

P(4 positive results) = .12^4

= .0002076 = 0.02076%

P(4 positives)=[P(1 positive)]^4= .12x.12x.12x.12 = .0002076

move the decimal to the right 2 places to get the percentage probability

= 0.02076%

or about

= 0.02% rounded off to 2 decimal places

if that doesn't "work" then you must have an answer key?

which may be wrong. It's happened before.

I think that's the PPPP, 4 Positives, you're referrring to, so is about 0.02% wrong according to an answer key, instructor or textbook?

what is the answer you think you're supposed to get?

or is this one of those Pearson or online type problems where they just tell you it's wrong with no clue what the right answer is supposed to be?

with Pearson and some similar online programs, you may have exactly the right answer, but the computer program rejects it due to some formatting requirements. That happens more often. You have to type it in exactly in the manner they specify. Or else it gets marked incorrect. sometimes you think you've formated it exactly as they seem to specify, but one tiny little difference you may not notice at 1st, and it gets marked wrong. Personally, Pearson seems to be the worst computer program witht he most "bugs" in it.

if the problem was 3 positives with P(1 positive) = .02

then just

find .02^3 = .000008 = .0008% probability

or maybe they want the answer in fractional form? that would = 1/125000

that's one of the "formatting" problems that might arise.

but you say to 4 decimal places, so probably not