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 can help me out to solve it?
I think the question is most likely not one tire but one car i.e what is the probability that at least one of the cars has no working tires?
Ans: case1: only one car has no working tires = 0.1^4(4C0*(0.9)^4+4C1*(0.9)^3(0.1)+4C1*(0.9)^2(0.1)^2+4C1*(0.9)^1(0.1)^3+4C1*(0.1)^4) = 1*0.1^4
As there two cars the probability is 2*0.1^4
But we are counting twice the probability of getting both the cars no working tires, so subtracting 0.1^4*0.1^4 from the above we get the following,
Probability = 19999/100000000 = 1.9999*10^-8
Hope this helps your query.