A person just bought new two cars ( 4 tires each). For each tire, the probabilit

Question

A person just bought new two cars ( 4 tires each). For each tire, the probability of being flattened is 10% . In other words, 90% of chance to be a regular, unfaulty tire. The person installs 4 tires on the first car and then install other 4 tires on the second car. (After finishing the installation on the first car) What would be the probability at least 1 tire out of 8 is flattened? I thought answer was 1(0.9 )^ 8, but my professor says it is wrong. Is someone in Chegg can solve this question?

Explanation / Answer

Given that First car's type have been installed. i.e. only four tyres are installed and remaining 4 are yet to install.

So we need to consider only four tyres.

Let p = probability of tyre being flattened is 0.10.

Hence Binomial (n=4,p=0.1), now using binomial probability distribution formula,

We need to find the probability P(X=0) = 1 - [(nCx)*(p^n)(1-p)^(n-x) ]= 1 - [(4c1)*(0.1^0)*(0.9^4)]

P(X=0) = 1 - 0.9^4 = 1 - 0.6561 = 0.3439

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