Imum Coeli
Junior Member
- Joined
- Dec 3, 2012
- Messages
- 86
Hi. Can anyone please help me with this question?
Q: Eleven taxis are dispached in such a way that four go to airport A, six got to airport B and one goes to airport C.
a) How many ways can this be accomplished?
b) If exactly one taxi is in need of repair what is the probability that it is dispached to airport C?
c) If exactly three of the taxis are in need of repair what is the probability that every airport recieves one of the taxis needing repair?
A:
a) There are 11 ways to choose taxi for airport C. There are 10 choose 6 ways to choose taxis for airport B. There is only one way to chose the remaining four taxis for airport A. By the multiplication rule (MN rule) then there are 2310 ways this can be achieved.
b) Since there are 11 cars to choose from to go to C and only one is defective the probability a taxi in need of repair being dispached to airport C is 1/11.
c) Suppose each airport has a taxi. Then
- there are 3! ways to distribute the broken taxis.
- 8 choose 0 ways to distribute good taxis to airport C
- 8 choose 5 ways to distribute good taxis to airport B
- 3 choose 3 ways to distribute good taxis to airport A
By MN rule there are 336 ways this can be done.
Therefore the probability of every airport recieving a broken taxi is 336/2310 = 8/55
Q: Eleven taxis are dispached in such a way that four go to airport A, six got to airport B and one goes to airport C.
a) How many ways can this be accomplished?
b) If exactly one taxi is in need of repair what is the probability that it is dispached to airport C?
c) If exactly three of the taxis are in need of repair what is the probability that every airport recieves one of the taxis needing repair?
A:
a) There are 11 ways to choose taxi for airport C. There are 10 choose 6 ways to choose taxis for airport B. There is only one way to chose the remaining four taxis for airport A. By the multiplication rule (MN rule) then there are 2310 ways this can be achieved.
b) Since there are 11 cars to choose from to go to C and only one is defective the probability a taxi in need of repair being dispached to airport C is 1/11.
c) Suppose each airport has a taxi. Then
- there are 3! ways to distribute the broken taxis.
- 8 choose 0 ways to distribute good taxis to airport C
- 8 choose 5 ways to distribute good taxis to airport B
- 3 choose 3 ways to distribute good taxis to airport A
By MN rule there are 336 ways this can be done.
Therefore the probability of every airport recieving a broken taxi is 336/2310 = 8/55