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Answer:

Probability at least one car will get punctured: 0.39347

Step-by-step explanation:

B(10,000 , 0.00005)

P(X ≥ 1) = 1 - P(X = 0)

            = 1 - (1 - 0.00005)^10,000

            = 1 - (0.99995)^10,000

            = 1 - 0.60652...

            = 0.39347 (probability that at least one car will get punctured)

As you can tell P(X ≥ 1) as we have to solve for the probability that at least one car will get punctured. That is of course 1 - [ P(X = 0) ].

How do we know to use a binomial distribution here? Based on these three reasons:

  1. Each event is independent. Each tire puncture is independent of any other puncture. The probability of puncture is constant at p = 0.00005 (there are four zeros between the decimal point and the 5).
  2. There are a fixed number of trials. In this case, there are n = 10,000 trials.
  3. There are two outcomes. Either a tire is punctured, or it is not punctured.
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