A test for the presence of a certain disease has probability
.20 of giving a false-positive reading (indicating that an
individual has the disease when this is not the case) and
probability .10 of giving a false-negative result. Suppose
that ten individuals are tested, five of whom have the
disease and five of whom do not. Let X 5 the number of
positive readings that result.
a. Does X have a binomial distribution? Explain your
reasoning.
b. What is the probability that exactly three of the ten
test results are positive?
The probabilities assigned here, for example, to the first case is not understandable since to me the prob(success) of having 0 detected out 5 affected people is supposed to be 0.1.
HINT
See the chart below with the probabilities for the four quadrants.
$\quad\quad\quad\quad\quad\quad\quad\quad$Test +$\quad\quad$ Test -
Have disease $\quad\quad\quad\; 0.9a\quad\quad\quad 0.1b$
Don't have disease $\quad 0.2c\quad\quad\quad\ 0.8d$
a = true positive, b = false negative
c = false positive, d = true negative
You should be able to proceed from here
Further Hint
It is given that $5$ are diseased and $5$ non-diseased, but only $3$ have been marked as positive.
In the first case, all the $5$ diseased have been marked false negative, and from the $5$ non-diseased, $3$ have been marked false positive, and $2$ as true negative, so the first row should be
$\binom5 0 0.9^0*0.1^5 \times \binom5 3 *0.2^3*0.8^2$
With this interpretation of D and D', the book answer seems awry !