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...

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.

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

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

To find the answers to these questions, we need to use the probabilities given for each type of vehicle. Let's go through each question step by step:

1. To find the probability that an individual would rent a sport utility vehicle (SUV) or a van (V), we need to add the probabilities of renting an SUV and a van together.

P(SUV or V) = P(SUV) + P(V) = 0.12 + 0.24 = 0.36

Therefore, the probability is 0.36.

2. To find the probability that an individual does not rent a midsize car (MC), we have to subtract the probability of renting a midsize car from 1 (since the sum of all probabilities should equal 1).

P(not MC) = 1 - P(MC) = 1 - 0.38 = 0.62

Therefore, the probability is 0.62.

3. To find the probability that none of the three individuals rents a midsize car (MC), we need to multiply the probabilities of each individual not renting a midsize car.

P(none MC) = P(not MC) x P(not MC) x P(not MC) = 0.62 x 0.62 x 0.62 ≈ 0.238328

Therefore, the probability is approximately 0.238328.

4. To find the probability that at least one of the three individuals would rent a compact car (CC), we need to find the probability of the complement event - the probability that none of the three individuals rent a compact car - and subtract it from 1.

P(at least one CC) = 1 - P(none CC)

To find P(none CC), we can use the multiplication rule. Since each individual is independent of the others, we multiply their individual probabilities of not renting a compact car.

P(none CC) = P(not CC) x P(not CC) x P(not CC) = (1 - P(CC)) x (1 - P(CC)) x (1 - P(CC))
= (1 - 0.16) x (1 - 0.16) x (1 - 0.16) ≈ 0.491776

Therefore, P(at least one CC) = 1 - 0.491776 ≈ 0.508224.

So, the probability is approximately 0.508224.

5. To find the probability that neither person rents a sport utility vehicle (SUV), we need to multiply the probabilities of each person not renting an SUV.

P(neither SUV) = P(not SUV) x P(not SUV)

P(not SUV) = 1 - P(SUV) = 1 - 0.12 = 0.88

P(neither SUV) = 0.88 x 0.88 = 0.7744

Therefore, the probability is 0.7744.

I hope this helps you understand how to find the answers to these questions. Let me know if you have any further questions!