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A manufacturer puts tires on two cars (four tires each). Each tire has a 10% probability of being flat. Assume the manufacturer puts four tires on one car first and then another four on the next car second. What’s the probability that there is at least one car that has no working tires?
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The probability that there is at least one car that has no working tires is given as follows:

0.0002 = 0.02%.

What is the binomial distribution formula?

The mass probability formula, giving the probability of x successes, is of:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are given by:

  • n is the number of trials of the experiment.
  • p is the probability of a success on a single trial of the experiment.

For each car, the parameters are given as follows:

p = 0.9, n = 4.

Hence the probability that a car has no working tires is given as follows:

P(X = 0) = (1 - 0.9)^4 = 0.0001.

Hence the probability that two cars have working tires is given as follows:

P(X = 2) = (1 - 0.0001)² = 0.9998.

The probability that at least one has no working tires is given as follows:

P(X < 2) = 1 - 0.9998 = 0.0002 = 0.02%.

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