Question

For the binomial distribution with n = 10 and p = 0.3, find the probability of: 1. Five or more successes. 2. At most two successes. 3. At least one success. 4. At least 50% successes

Answer 1

Solution:
Given in the question
No. of sample = 10
Probability of success P(Success) = 0.3
binomial distribution formula is
P(X=x | n) = nCx*(p)^x * (1-p)^(n-x)
Solution(1)
P(X>=5) = P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) = 10C5*(0.3)^5*(0.7)^5 +10C6*(0.3)^6*(0.7)^4 +10C7*(0.3)^7*(0.7)^3 +10C8*(0.3)^8*(0.7)^2 +10C9*(0.3)^9*(0.7)^1 +10C10*(0.3)^10*(0.7)^0 = 0.1503
Solution(2)
P(X<=2) = P(X=0) + P(X=1) + P(X=2) = 10C0*(0.3)^0*(0.7)^10 +10C1*(0.3)^1*(0.7)^9 +10C2*(0.3)^2*(0.7)^8 = 0.3828
Solution(3)
P(X>=1) = 1-P(X<1) = 1 - P(X=0) = 1- 10C0*(0.3)^0*(0.7)^10 = 1-0.0282 = 0.9718
Solution(4)
At least 50% Success means Success should be 5 or more
P(X>=5) = P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) = 10C5*(0.3)^5*(0.7)^5 +10C6*(0.3)^6*(0.7)^4 +10C7*(0.3)^7*(0.7)^3 +10C8*(0.3)^8*(0.7)^2 +10C9*(0.3)^9*(0.7)^1 +10C10*(0.3)^10*(0.7)^0 = 0.1503