Drug Trials: prob. that 4 of 15 will be cured? 3 to 6? etc

A drug firm administers a new drug to 15 people with a certain disease. If the probability is 0.20 that the drug will cure each person of the disease:
a) What is the probability that 4 of the 15 people will be cured?
b) What is the probability that between 3 and 6 (inclusive) of the people will be cured?
c) What is the probability that fewer than 5 of the people will be cured?
d) What is the probability that none of the people will be cured?
e) What is the probability that all of the people will be cured?

This involves the binomial probability distribution formula:

p(r) = nCr * p^r * q^(n-r)

r = number of outcomes of interest
n = number of trials
nCr = n! / (r!*(n-r)!)
p = probability of success
q = 1-p

(a) r=4, so:
p(4) = 15C4 * 0.2^4 * 0.8^11

(b) r = 3,4,5,6
p(3) = 15C3 * 0.2^3 * 0.8^12
p(4) = see problem a
p(5) = 15C5 * 0.2^5 * 0.8^10
p(6) = 15C6 * 0.2^6 * 0.8^9
then add up these 4 answers

(c) r = 0,1,2,3,4

(d) r = 0

(e) r = 15