# Test for a certain disease

• Oct 19th 2010, 05:24 PM
mightydog78
Test for a certain disease
Testing for a disease. has 0.02 probability of giving a false-positive reading and probability of 0.03 of giving a false-negative result. (a false-positive means testing postive for the disease when it's a healthy person).
5 people are test, two of whom have the disease and three who don't. X=number of postive readings.

I am going to assume this is a binomial distribution and there are only 2 outcomes (postive or negative for disease). Right?

What im having trouble with is the probability that at most 2 of the 5 tests results are positive.. I am confused where to start this (Nerd)
• Oct 19th 2010, 11:53 PM
mr fantastic
Quote:

Originally Posted by mightydog78
Testing for a disease. has 0.02 probability of giving a false-positive reading and probability of 0.03 of giving a false-negative result. (a false-positive means testing postive for the disease when it's a healthy person).
5 people are test, two of whom have the disease and three who don't. X=number of postive readings.

I am going to assume this is a binomial distribution and there are only 2 outcomes (postive or negative for disease). Right?

What im having trouble with is the probability that at most 2 of the 5 tests results are positive.. I am confused where to start this (Nerd)

X ~ Binomial(n = 5, p = ?)

So your first job is to calculate the value of p, the probability of a positive result in a single trial.

Then calculate $\Pr(X \leq 2)$.
• Oct 20th 2010, 09:00 PM
mightydog78
P(randomly chosen person gives + test) = 0.4*0.97 + 0.6*0.02 = 0.4 so is this my p value??

P(X≤2) = P(X=0) + P(X=1) + P(X=2) = (0.6)⁵+ 5C1*(0.4)*(0.6)⁴+ 5C2*(0.4)²*(0.6)³= 0.6826

and then do i go to the binomial tables and find where 0.6826 and 0.4 intersect??
• Oct 20th 2010, 09:04 PM
mightydog78
wow im dumb, i calculated the probability haha