A machine produces parts of which 0.2% are defective. If a random sample of ten parts produced by this machine contains two or more defectives, the machine is shut down for repairs. Find the probability that the machine will be shut down for repairs based on this sampling plan.

Probability= 0.199995904

Step-by-step explanation:

Probabilty of defective p = 0.2

Probability of non defective q = 1-0.2

= 0.8

Probability that 2 or more will be defective out of ten = probability of all defective - probability of 1 non defective

Probability of1 non defective

= 10C1(0.8)¹(0.2)^9

= 10*0.8*5.12*10^-7

= 4.096*10^-6

= 0.000004096

Probability of two or more defective

= 0.2-0.000004096

= 0.199995904

the solution is h=10 because you add 9 to each side of the equation.