At a car rental agency, 0.34 of the cars are returned on time. A sample of 13 car rentals is studied. What is the probability that more than 3 of them are returned on time?
Write only a number as your answer. Round to 2 decimal places ( for example: 0.24). Do not write as a percentage.
prob(ontime) = .34
prob(not ontime) = (.66)
more than 3 of 13 implies
we CAN'T have 0 return + 1 return + 2 return + 3 return
so prob(more than 3 return) = 1 - (sum of the 4 cases listed)
I will do one of the 4 cases, you do the rest
prob(3 of the 13 are returned)
= C(13,3)(.34)^3 (.66)^10
= 286(.039304)(.015884)
= appr .1763
carry about 4 decimals since you want only 2 in your answer
To find the probability that more than 3 of the cars are returned on time, we need to calculate the cumulative probability of 0, 1, 2, and 3 correctly returned cars and subtract it from 1.
First, let's find the probability of each scenario:
Probability of 0 out of 13 cars returned on time:
P(X = 0) = (1 - 0.34)^13
Probability of 1 out of 13 cars returned on time:
P(X = 1) = 13C1 * (0.34^1) * (1 - 0.34)^(13-1)
Probability of 2 out of 13 cars returned on time:
P(X = 2) = 13C2 * (0.34^2) * (1 - 0.34)^(13-2)
Probability of 3 out of 13 cars returned on time:
P(X = 3) = 13C3 * (0.34^3) * (1 - 0.34)^(13-3)
To find the cumulative probability, we add up the probabilities of each scenario mentioned above:
Cumulative probability of more than 3 cars returned on time:
P(X > 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))
Now we can calculate it:
P(X > 3) = 1 - [(1 - 0.34)^13 + 13C1 * (0.34^1) * (1 - 0.34)^(13-1) + 13C2 * (0.34^2) * (1 - 0.34)^(13-2) + 13C3 * (0.34^3) * (1 - 0.34)^(13-3)]
By evaluating the expression, we get the probability that more than 3 cars are returned on time.