Answer:
0.1979 or 19.79%
Step-by-step explanation:
If 20% of all bulbs are defective, there is a 20% chance of each bulb being defective and an 80% chance of each bulb not being defective.
This is a binomial probability model with probability of success (bulb being defective) of p=0.20.
In order for the lot to pass inspection, it must contain either zero or one defective bulb, the probability of one of these scenarios occurring is:
[tex]Pass= P(d=0)+P(d=1)\\Pass= 0.80^{14}+14*0.20*0.80^{13}\\Pass=0.1979[/tex]
The probability that the lot will pass inspection is 0.1979 or 19.79%.
