The Reindeer Shuttle has 2 vans taking travelers from West Lafayette to Chicago’
ID: 3272468 • Letter: T
Question
The Reindeer Shuttle has 2 vans taking travelers from West Lafayette to Chicago’s O’Hare airport. The large shuttle can hold up to 9 passengers and the small shuttle can hold up to 6 passengers. On a particular day both vans travel to Chicago with 1 or more passengers. Assume that all possible outcomes are equally likely. (Note: you may want to sketch the sample space for yourself—using ordered pairs like (L, S) where the first number represents the number of passengers in the large van and the second represents the number of passengers in the small van) e.g. (8, 3).)
What is the probability of the following events:
A: Together the 2 vans carry a total of 4 or 6 or 10 passengers.
B: The large van carries twice as many passengers as the small van.
C: The two vans carry different numbers of passengers.
Explanation / Answer
a) As the large van can carry any number of passengers from 1 to 9 , therefore there are 9 possibilities for the number of passengers in the large van. The small shuttle can carry any number of passengers from 1 to 6, therefore there are 6 possibilities for the small shuttle. Therefore the total number of combinations for the 2 vehicles would be computed as: 9*6 = 54 combinations that is there would be 54 outcomes in the sample space. 6 outcomes for each of the 9 outcomes.
a) Together they carry a total of 4 or 6 or 8 passengers. This could be done in the following cases:
Therefore there are a total of 3 +5 + 6 = 14 cases when this can happen. Therefore the required probability here would be:
= 14/54 = 0.2593
Therefore 0.2593 is the required probability here.
b) The large van carries twice as many passengers as the small van. This can happen in the following ways: 2 + 1, 4 + 2, 6 + 3, 8 + 4. Therefore there are 4 ways in which this is possible. Hence the required probability is:
= 4/54 = 0.0741
Therefore 0.0741 is the required probability here.
c) The 2 vans carry different number of passengers.
= 1 - Probability that the 2 vans carry same number of passengers.
2 vans can carry same number of passengers in the following was: 1 + 1, 2+ 2, 3+ 3, 4 + 4, 5+ 5, 6+ 6 that is in 6 ways. Therefore they carry different number of passengers in 54- 6 = 48 ways. Therefore the required probability here would be:
= 48/54 = 0.8889
Therefore 0.8889 is the required probability here.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.