Intro to Statistics Help!!!?
March 17th, 2010 | by Rarzi |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
Consider the next four individuals who rent cars. (Assume they come in randomly and do not know each other.) Find the probability that none of the four rent a van. Show your results as a decimal rounded to six decimal places; do not show your answer as a percent.
Answer ______
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
Consider the next four individuals who rent cars. (Assume they come in randomly and do not know each other.) Find the probability that at least one of the four rent a van. Show your results as a decimal rounded to six decimal places; do not show your answer as a percent.
Answer ___________________
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
Consider the next three individuals who rent cars. (Assume they come in randomly and do not know each other.) Find the probability that all three individuals would rent a sport utility vehicle (SUV). Show your answer to 6 decimal places. Do not show the answer as a percent.
__________________
One Response to “Intro to Statistics Help!!!?”
By amgasiago on Mar 17, 2010 | Reply
assuming each person’s decision is independent of the decisions of all other people:
to solve for the probably of non of the four renting a van, first find the probability of someone renting a van, which is .24, and the probability of them not renting a van is 1-.24, which is .76, so to get the probability of all four people not renting a van, put .76^4 = (.76)(.76)(.76)(.76)
to solve for the probability that at least one of the four rents a van, take 1-P(none of the renting a van) = 1-(your previous answer) = 1-[(.76)(.76)(.76)(.76)]
to solve for the probability that all three rent an SUV, take the probility that one rents an SUV, .12 and put that to the third power, or (.12)(.12)(.12)