Question

On a particular production line, the likelihood that a light bulb is defective is 14%. six light bulbs are randomly selected. What is the probability that at most 4 of the light bulbs will be defective?

Answer 1

Solution

Given that ,

p = 0.14

1 - p = 0.86

n = 6

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * px * (1 - p)n - x

P(X 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= ((6! / 0! (6 - 0)!) * 0.140 * (0.86)6 - 0 + ((6! / 1! (6 - 1)!) * 0.141 * (0.86)6 - 1 + ((6! / 2! (6 - 2)!) * 0.142 * (0.86)6 - 2​​​​​​​ + ((6! / 3! (6 - 3)!) * 0.143 * (0.86)6 - 3​​​​​​​ + ((6! / 4! (6 - 4)!) * 0.144 * (0.86)6 - 4

=  0.4046 + 0.3952 + 0.1608 + 0.0349 + 0.0043

Probability = 0.9997