Question

1. a car dealership leases cars to customers. the table shows the number of leases signed for each type of car on the lot for the 100 most recent contracts. during a seasonal audit, the manager randomly selects 3 lease contracts to proofread without replacing them.

type of car number of leases
compact 45
sedan 32
convertible 7
suv 16

a. what is the probability that the manager selects a lease contract for a convertible first?

b. what is the probability that the manager selects a contract for an suv first and a contract that is not for a compact car second?

c. what is the probability that all 3 of the contracts that the manager selects are for compact cars?

d. what is the probability that none of the contracts that the manager selects are for compact cars?

Explanation:

Step1: Recall probability formula

The probability of an event $P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. The total number of lease - contracts is $n = 45+32 + 7+16=100$.

Step2: Calculate probability for part a

For the first - contract selection, the number of convertible contracts is $n_{convertible}=7$. So the probability that the first contract is for a convertible is $P(\text{convertible first})=\frac{7}{100}=0.07$.

Step3: Calculate probability for part b

The probability that the first contract is for an SUV is $P(\text{SUV first})=\frac{16}{100}$. After selecting an SUV first (without replacement), the number of non - compact contracts left is $(32 + 7+16)-1=54$ (since one SUV is already selected) and the total number of contracts left is $99$. So the probability that the second contract is not for a compact car given that the first is for an SUV is $P(\text{not compact second}|\text{SUV first})=\frac{54}{99}$. By the multiplication rule of probability for dependent events $P(A\cap B)=P(A)\times P(B|A)$, the probability that the manager selects a contract for an SUV first and a contract that is not for a compact car second is $\frac{16}{100}\times\frac{54}{99}=\frac{16\times54}{100\times99}=\frac{864}{9900}\approx0.0873$.

Step4: Calculate probability for part c

The probability that the first contract is for a compact car is $P(\text{compact first})=\frac{45}{100}$. After selecting a compact car first (without replacement), the number of compact cars left is $44$ and the total number of contracts left is $99$. The probability that the second contract is for a compact car given that the first is for a compact car is $P(\text{compact second}|\text{compact first})=\frac{44}{99}$. After selecting two compact cars, the number of compact cars left is $43$ and the total number of contracts left is $98$. The probability that the third contract is for a compact car given that the first two are for compact cars is $P(\text{compact third}|\text{compact first and second})=\frac{43}{98}$. By the multiplication rule of probability for dependent events, $P(\text{all 3 compact})=\frac{45}{100}\times\frac{44}{99}\times\frac{43}{98}=\frac{45\times44\times43}{100\times99\times98}=\frac{85140}{970200}\approx0.0877$.

Step5: Calculate probability for part d

The number of non - compact contracts is $32 + 7+16 = 55$. The probability that the first contract is not for a compact car is $P(\text{not compact first})=\frac{55}{100}$. After selecting a non - compact car first (without replacement), the number of non - compact cars left is $54$ and the total number of contracts left is $99$. The probability that the second contract is not for a compact car given that the first is not for a compact car is $P(\text{not compact second}|\text{not compact first})=\frac{54}{99}$. After selecting two non - compact cars, the number of non - compact cars left is $53$ and the total number of contracts left is $98$. The probability that the third contract is not for a compact car given that the first two are not for compact cars is $P(\text{not compact third}|\text{not compact first and second})=\frac{53}{98}$. By the multiplication rule of probability for dependent events, $P(\text{none compact})=\frac{55}{100}\times\frac{54}{99}\times\frac{53}{98}=\frac{55\times54\times53}{100\times99\times98}=\frac{155910}{970200}\approx0.1607$.

Answer:

a. $0.07$
b. $\frac{864}{9900}\approx0.0873$
c. $\frac{45\times44\times43}{100\times99\times98}\approx0.0877$
d. $\frac{55\times54\times53}{100\times99\times98}\approx0.1607$