Question

A test for the presence of a certain disease has probability 0.2 of giving a false-positive result (indicating that an individual has the disease when this is not the case) and probability 0.1 of giving a false-negative result. Suppose that 10 individuals are tested, 5 of whom have the disease and 5 whom do not. Let X be the number of positive readings that result.

a. Explain why X does not have a binomial distribution.

b. find the probability that exactly 3 of the 10 test results is positive?

Answer 1

baye thrm, yay!!

Answer 2

or am i off on that

Answer 3

Poisson, I think. Not sure, but would make sense since it's in the chapter

Answer 4

prior probability is .5 yes .5 no


.5(.2)
------------ = probability of yes when no if i see this
.5(.2) + .5(.1) part right

Answer 5

or should .1 be .9

Answer 6

I'm supposed to get .0273 for part b :/

Answer 7

.1 is a no when yes; we want the yes with yes so id say its .9 lets try that and see where it leads me

Answer 8

.5(.2)
------------ = probability of yes when no = 10/55
.5(.2) + .5(.9)


.5(.9)
------------ = probability of yes when yes = 45/55
.5(.2) + .5(.9)

would that look right?

Answer 9

I'm not sure, that's the problem. I tried using poisson but no luck

Answer 10

if we add them we get a number greater than one which is why im wondering it this is even part of the right track

P(test+) = P( +|y or +|n) ? but if we and it we get:

450/3025 = .1488 for a + result on any given person hmmm

Answer 11

keep in mind that i am trying to come to a result based on what I know; and that aint alot :)

Answer 12

40/45 * 5/45 = 200/2025 = .0988 for a - result on any given person

120 *.1488^3 *.0988^7 = .. not even close to the answer you gave

Answer 13

\[\sum_{x=0}^{3}{5\choose x}{5\choose 3-x}(.2)^{x}(.9)^{3-x}(.8)^{5-x}(.1)^{x+2}\]

Answer 14

\[=\frac{53317}{1953125}\]