Question

The probability of a defective unit in a manufacturing process has a binomial distribution with p = 0.03 What is the probability of 2 or more occurences in a sample of 15 units?

(Hint: Probability of 2 or more is 1 minus probability of 1 or less.)

Answer 1

Answer: Let X be the random variable denoting the number of defective units in a manufacturing process.

Given X ~ Binomial(15,0.03)

To find P(X >= 2) = 1 - P(X <=1) = 1 - [ P(X = 0) + P(X=1)]

= 1 - [((15C0)*((0.03)⁰)*(1-0.03)¹⁵) + (15C1)*((0.03)¹)*(1-0.03)¹⁴)]

= 0.07297249

= 0.073 (Rounded to three decimal places)

Since the probability mass function of binomial distribution is P(x; p,n) = (nCx) * (p^x)*(1-p)^n-x., for x = 0,1,2,...,n

where nCx = factorial(n) / (factorial(x) * factorial(n-x))

Here n = 15, p 0.03