Question
Ricardo's car rental offers five different types of vehicles, compact cars (CC), midsize cars (MC), sport utility vehicles (SUV), vans (V), and luxury cars (L).
Historically, if a random individual comes to rent a vehicle, the following probabilities apply:
P(CC) = 0.16
P(MC) = 0.38
P(SUV) = 0.12
P(V) = 0.24
1. Find the probability that an individual would rent a sport utility vehicle (SUV) or a van (V). Show your answer as a decimal. Do not show the answer as a probability.
2. Find the probability that an individual does not rent a midsize car (MC). Show your answer as a decimal; do not show your answer as a percent.
3. Consider the next three individuals who rent cars. (Assume they come in randomly and do not know each other.) Find the probability that none of the three rents a midsize car (MC). Show your results as a decimal rounded to six decimal places; do not show your answer as a percent.
4. Consider the next three individuals who rent cars. (Assume they come in randomly and do not know each other.) Find the probability that at least one of the three individuals would rents a compact car (CC). Show your answer to 6 decimal places. Do not show the answer as a percent.
5. Two people arrive to rent cars. What is the probability that neither person rents a sport utility van (SUV)? Show your answer as a decimal with four decimal places; do not show your answer as a percent.
I am lost on how to find the answer to these questions...
Historically, if a random individual comes to rent a vehicle, the following probabilities apply:
P(CC) = 0.16
P(MC) = 0.38
P(SUV) = 0.12
P(V) = 0.24
1. Find the probability that an individual would rent a sport utility vehicle (SUV) or a van (V). Show your answer as a decimal. Do not show the answer as a probability.
2. Find the probability that an individual does not rent a midsize car (MC). Show your answer as a decimal; do not show your answer as a percent.
3. Consider the next three individuals who rent cars. (Assume they come in randomly and do not know each other.) Find the probability that none of the three rents a midsize car (MC). Show your results as a decimal rounded to six decimal places; do not show your answer as a percent.
4. Consider the next three individuals who rent cars. (Assume they come in randomly and do not know each other.) Find the probability that at least one of the three individuals would rents a compact car (CC). Show your answer to 6 decimal places. Do not show the answer as a percent.
5. Two people arrive to rent cars. What is the probability that neither person rents a sport utility van (SUV)? Show your answer as a decimal with four decimal places; do not show your answer as a percent.
I am lost on how to find the answer to these questions...
Answers
Probabilities only add to .90. Is the remaining .10 people who decided not to rent, or do you have a typo?
1. Either-or probabilities are found by adding the individual probabilities.
2. .16 + .12 + .24 = ?
3. If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
(answer to 2)^3 = ?
4. At least one = one, two or all rent CC. Use principles above.
5. Use principles above.
1. Either-or probabilities are found by adding the individual probabilities.
2. .16 + .12 + .24 = ?
3. If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
(answer to 2)^3 = ?
4. At least one = one, two or all rent CC. Use principles above.
5. Use principles above.
No typo, that is how the question reads.
Can you check my answers?
1. 0.54
2. 0.52
3. 0.140608
4. 0.405224
5. 0.4746
Can you check my answers?
1. 0.54
2. 0.52
3. 0.140608
4. 0.405224
5. 0.4746
1. SUV = .12, V = .24.
.12 + .24 ≠ .54
2. right
3. right, however, there is a problem with the missing .10. If it is included, it would be (1-.38)^3. = .62^3 = .388244
4. .16 + .16^2 + .16^3 = .238328
5. again to include the .10, (1-.12)^2 = .88^2 = .7744
.12 + .24 ≠ .54
2. right
3. right, however, there is a problem with the missing .10. If it is included, it would be (1-.38)^3. = .62^3 = .388244
4. .16 + .16^2 + .16^3 = .238328
5. again to include the .10, (1-.12)^2 = .88^2 = .7744
Related Questions
Assume that you are to pick 3 cars from the motor pool, which contains 6 subcompact cars, 6 compact...
Suppose the National Transportation Safety Board (NTSB) wants to examine the safety of compact cars,...
A car rental agency rents compact, midsize, and luxury cars. Its goal is to purchase 60 cars with a...