Use the binomial expression (p+q)n to calculate a binomial distribution with n = 5 and p = 0.3.
The binomial distribution formula is:
P(X=k) = C(n,k) * p^k * q^(n-k)
where C(n,k) is the number of ways to choose k successes out of n trials and is calculated as C(n,k) = n! / (k! * (n-k)!).
In this case, n = 5, p = 0.3, and q = 0.7 (since q = 1 - p).
Let's calculate the binomial distribution for k=0, 1, 2, 3, 4, 5:
P(X=0) = C(5,0) * (0.3)^0 * (0.7)^5 = 1 * 1 * 0.16807 = 0.16807
P(X=1) = C(5,1) * (0.3)^1 * (0.7)^4 = 5 * 0.3 * 0.2401 = 0.36015
P(X=2) = C(5,2) * (0.3)^2 * (0.7)^3 = 10 * 0.09 * 0.343 = 0.3087
P(X=3) = C(5,3) * (0.3)^3 * (0.7)^2 = 10 * 0.027 * 0.49 = 0.1323
P(X=4) = C(5,4) * (0.3)^4 * (0.7)^1 = 5 * 0.0081 * 0.7 = 0.02835
P(X=5) = C(5,5) * (0.3)^5 * (0.7)^0 = 1 * 0.00243 * 1 = 0.00243
So, the binomial distribution for n=5 and p=0.3 is:
P(X=0) = 0.16807
P(X=1) = 0.36015
P(X=2) = 0.3087
P(X=3) = 0.1323
P(X=4) = 0.02835
P(X=5) = 0.00243