Dividing students to three Buses 5 friends want to join a school trip. Students attending the trip are shared randomly on 3 buses, A, B and C.


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*What is the probability that the all the 5 friends are on bus A?

*What is the probability that the all the 5 friends are on any one of the buses?

*What is the probability that 3 of the 5 friends are on any one of the buses?


I tried solving but in terms of probability, we should divide by the total number of ways, but I couldn't get that since there is no total number of students to choose from neither is there a number of students in each bus/group.
Any idea?
 A: The underlying assumption is that you can fit as many (or few) of the five friends on any of the buses, and the rest of the students (which you don't care about for the purposes of this problem) will adjust to accommodate.
Each of the five friends can be on any of the buses, so the total number of possibilities is $3^5$. Then it's a matter of counting the number of possibilities for each case.


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*They're all on Bus A, so there's only one choice.

*Choose the bus they're all on. There are three buses, so therefore three choices.

*Pick the three friends on the same bus $(_5C_3)$ and pick the bus $(3)$. Then pick the buses that the other two friends are on $(2^2)$.


If #3 involves at least three on the same bus, then repeat for four friends on the same bus, and then five (which you've already done).
Expanding on my answer to number 3: We need to multiply those three numbers together to get all of the possibilities. We can choose the three friends on the same bus, but there are three choices for which bus any particular three friends are on. Once we have that, there are $2^2=4$ choices for where the two left-out friends go, given any particular group of three friends and any particular bus.
By the time we're done, the probability is $120/243$, or nearly one-half.
