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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?

Mathematics
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No, that wasn't the answer. This question is in a section over Bernoulli Trials.
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
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yes
Mubzz thank you yes!!
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!!

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