I need help with this question.Warranty records show that the probability that a new car needs a warranty repair in the first 90 days is 0.05.  If a sample of 3 new cars is selected:What is the probability that none needs a warranty repair? __________________What is the probability that at least one needs a warranty repair? _________________What is the probability that more than one needs a warranty repair? __________________

Question
Asked Nov 4, 2019

I need help with this question.

Warranty records show that the probability that a new car needs a warranty repair in the first 90 days is 0.05.  If a sample of 3 new cars is selected:

  1. What is the probability that none needs a warranty repair? __________________
  2. What is the probability that at least one needs a warranty repair? _________________
  3. What is the probability that more than one needs a warranty repair? __________________
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Expert Answer

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Step 1

The probability that none needs a warranty repair is obtained below:

Binomial distribution:

The probability mass function (pmf) of a binomial random variable X is given as:

p*q*, x=0,1..,n;0 < p< 1;q = 1-p
P(x)
otherwise
0,
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p*q*, x=0,1..,n;0 < p< 1;q = 1-p P(x) otherwise 0,

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Step 2

Calculating the probability value:

The random variable X follows binomial distribution. Here, n= 3, x= 0, and p=0.05. The required probability is as follows:

P(x-0)-(0.0s(-003
0.05) (0.95'
1x1x 0.8574
0.8574
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P(x-0)-(0.0s(-003 0.05) (0.95' 1x1x 0.8574 0.8574

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Step 3

The probability that at least one needs a warranty repair is o...

P(x21)1-P(X 0)
3
=1-
os)(1-0.05)
3-0
1-(1)x (1)x0.8574
1-0.8574
0.1426
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P(x21)1-P(X 0) 3 =1- os)(1-0.05) 3-0 1-(1)x (1)x0.8574 1-0.8574 0.1426

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