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?

Answer 1

Let X be a number of defective units among 15 units

Here, X ~ Binomial ( n = 15, p = 0.03)

probability mass function of X is,

P(X = x) = ⁿC_x p^x ( 1 - p )^n-x

We want to find, P(X >= 2)

P(X >= 2)

= 1 - 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 ]

= 1 - [ 0.63325 + 0.29378 ]

= 1 - 0.92703

= 0.07297

P(X >= 2) = 0.07297

The probability is 0.07297