Warranty records show that that the probability that a new Mercedaz car needs a repair withinthe first 90 days after delivery is 5%.Find the probability that in a random sample of six new Mercedaz cars:(i) None will require a repair within the first 90 days after delivery(ii) At most one car will require a repair within the first 90 days after delivery
Question
Warranty records show that that the probability that a new Mercedaz car needs a repair withinthe first 90 days after delivery is 5%.Find the probability that in a random sample of six new Mercedaz cars:(i) None will require a repair within the first 90 days after delivery(ii) At most one car will require a repair within the first 90 days after delivery
Solution
This is a binomial distribution problem.
(i) The probability that none of the six cars will require a repair within the first 90 days after delivery can be calculated using the formula for binomial distribution:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability we want to find,
- C(n, k) is the number of combinations of n items taken k at a time,
- p is the probability of a single event (in this case, the probability that a car needs a repair within the first 90 days, which is 5% or 0.05),
- n is the number of trials (in this case, the number of cars, which is 6), and
- k is the number of successful events we want (in this case, the number of cars that need a repair, which is 0 for this part of the question).
So, P(X=0) = C(6, 0) * (0.05^0) * ((1-0.05)^(6-0)) = 1 * 1 * (0.95^6) = 0.7351 or 73.51%.
(ii) The probability that at most one car will require a repair within the first 90 days after delivery is the sum of the probabilities that exactly 0 cars and exactly 1 car will need a repair.
So, P(X=1) = C(6, 1) * (0.05^1) * ((1-0.05)^(6-1)) = 6 * 0.05 * (0.95^5) = 0.2219 or 22.19%.
Therefore, P(X<=1) = P(X=0) + P(X=1) = 73.51% + 22.19% = 95.7%.
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