p(x) = p^x * q^(n-x) * c(n,x)
when n = 4, x will range from 0 to 4
p(x) is the probability that x events will occur.
p is the probability that one event will occur.
q is the probability that one event will not occur.
p^x is p raised to the power of x.
q^(n-x) is q raised to the power of (n-x)
q is equal to 1 - p.
c(n,x) is the combination formula for obtaining the number of sets of x elements each out of the pool of n elements when order is not important.
the formula for c(n,x) is c(n,x) = n! / (x! * (n-x)!)
in your problem:
n is equal to 4
x goes from 0 to 4
p is equal to .4
q is equal to .6
the questions is what is the probability that the drug will be effective on 2 out of 4 patients.
that would be p(2) = .4^2 * .6^2 * c(4,2)
that becomes p(2) = .16 * .36 * 6
the result is p(2) = .3456
that's your solution.
the sum of all probabilities has to be equal to 1.
in this problem, that means that p(0) + p(1) + p(2) + p(3) + p(4) must be equal to 1.
i did a check through the use of excel and found that this is true.
if you find p(0) through p(4) and add them up, they will be equl to 1.
these are the numbers:
x p(x)
0 0.1296
1 0.3456
2 0.3456
3 0.1536
4 0.0256
sum 1.0000
p(1) and p(2) are the same probability, but that's not an error.
p(1) = .4^1 * .6^3 * 4! / (1! * 3!) = .4 * .216 * (4*3*2*1) / (1*3*2*1) = .4 * .216 * 4 = .3456
p(2) = .4^2 * .6^2 * 4! / (2! * 2!) = .16 * .36 * (4*3*2*1) / (2*1*2*1) = .16 * .36 * 6 = .3456
the overall probability of each is the same, but the elements within each overall probability are different.