Assume that 1 car in 8 is defective and will not start. If at different times 5 individuals each randomly select a car to test drive, what is the probability that at least 1 of them selects a car that will not start?

Question

Assume that 1 car in 8 is defective and will not start.  If at different times 5 individuals each randomly select a car to test drive, what is the probability that at least 1 of them selects a car that will not start?

anonymous · Answer

No, that wasn't the answer.  This question is in a section over Bernoulli Trials.

anonymous · Answer

The collective probability of any event cannot exceed 1. At least one of them selects a defected car means that there can be more than one defected cars picked.

So,
probability of at least 1 defective car = 1 - (probability of no defective cars)

For one person's pick, probability of non-defective car = 7/8.
For five different draws, this will be (7/8)^5

Therefore,

=>probability of at least 1 defective car = 1 - ((7/8)^5)
=>probability of at least 1 defective car = 1 - (16807/32768)
=>probability of at least 1 defective car = 1 - 0.5129
=>probability of at least 1 defective car = 0.4871

anonymous · Answer

Is that the correct answer?

zarkon · Answer

yes

anonymous · Answer

Mubzz thank you yes!!

anonymous · Answer

I thought it was somewhere along that line but i didnt subtract (7/8)^5 from 1.  Thats why i didnt get it thank you again!!