Industry standards suggest that 18% of new vehicles require warranty service within the first year. Jones Nissan, sold 11 Nissans yesterday
What is the probability that less than three of these vehicles require warranty service?
sorry
P(0 broke) = 1* 1 * .82^11
P(1 broke) = C(11,1).18^1 .82^10
P(2 broke) =C(11,2).18^2 .82^9
To determine the probability that less than three of the vehicles sold yesterday at Jones Nissan require warranty service, we can use the binomial probability formula.
The binomial probability formula is:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of getting exactly k successes
- n is the total number of trials (in this case, the number of vehicles sold)
- k is the number of successes we're interested in (in this case, the number of vehicles requiring warranty service)
- p is the probability of success on each trial (in this case, the probability of a vehicle requiring warranty service)
Given that industry standards suggest that 18% of new vehicles require warranty service within the first year, the probability of a vehicle requiring warranty service is p = 0.18.
Since we want to find the probability that less than three vehicles require warranty service, we need to calculate the probabilities for k = 0, 1, and 2, and then sum them up.
Let's calculate these probabilities using the formula:
P(X = 0) = C(n, 0) * p^0 * (1 - p)^(n - 0)
P(X = 1) = C(n, 1) * p^1 * (1 - p)^(n - 1)
P(X = 2) = C(n, 2) * p^2 * (1 - p)^(n - 2)
For each calculation, C(n, k) represents the number of combinations. In this case, it would be C(11, 0), C(11, 1), and C(11, 2), respectively.
Let's calculate these probabilities.
P(X = 0) = C(11, 0) * 0.18^0 * (1 - 0.18)^(11 - 0)
= 1 * 1 * (0.82)^11
≈ 0.1072
P(X = 1) = C(11, 1) * 0.18^1 * (1 - 0.18)^(11 - 1)
= 11 * 0.18 * 0.82^10
≈ 0.2837
P(X = 2) = C(11, 2) * 0.18^2 * (1 - 0.18)^(11 - 2)
= 55 * 0.18^2 * 0.82^9
≈ 0.3131
Finally, we can sum up these probabilities to get the probability that less than three vehicles require warranty service:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
≈ 0.1072 + 0.2837 + 0.3131
≈ 0.7040
Therefore, the probability that less than three of the vehicles sold yesterday at Jones Nissan require warranty service is approximately 0.7040 or 70.40%.
p yes broke = .18
p not broke = .82
P(0 broke) = 1*.18^11 * 1
P(1 broke) = C(11,1).18^10 .82^1
P(2 broke) =C(11,2).18^9 .82^2
C(11,1) = 11
C(11,2) = 55