Consider the different combinations of numbers of passengers that can get off at the four stops; the numbers must be whole numbers no larger than 8. Then consider the number of different ways each of those combinations can be distributed among the four stops.
combination # ways
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0 0 0 8 4
0 0 1 7 12
0 0 2 6 12
0 0 3 5 12
0 0 4 4 6
0 1 1 6 12
0 1 2 5 24
0 1 3 4 24
0 2 2 4 12
0 2 3 3 12
1 1 1 5 4
1 1 2 4 12
1 1 3 3 6
1 2 2 3 12
2 2 2 2 1
The total # of ways with 0 passengers getting off at 3 of 4 stops is 4
The total # of ways with 0 passengers getting off at 2 of 4 stops is 12+12+12+6=42
The total # of ways with 0 passengers getting off at 1 of 4 stops is 12+24+24+12+12=84
The total # of ways with 0 passengers getting off at 0 of 4 stops is 4+12+6+12+1=35
The total # of ways the 8 passengers can get off at the 4 stops is 4+42+84+35=165
Now for the specific questions that are asked....
(1) What is the probability of none of the passengers getting off on at least one bus stop?
This means the number of stops with no passengers getting off is 0 -- i.e., at least one passenger gets off at each stop. The probability is 35/165
ANSWER: 35/165
Simplify if desired or required
(2) And what is the probability of them not getting off on exactly 2 bus stops?
That is clearly the probability that the numbers of passengers getting off is 0 at exactly 2 of the 4 stops. The probability is 42/165.
ANSWER: 42/165
Again simplify if desired or required
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Note the problem can be worked using "stars and bars". If you are not familiar with that method and don't understand the discussion below, search this site or the internet for further information and examples.
The total number of ways of distributing the 8 passengers among the 4 stops without restrictions can be viewed as the number of different arrangements of 8 stars (representing the passengers) and 4-1=3 bars (representing the dividers that divide the 8 passengers into 4 groups). That number is
.
That agrees with the number we came up with the analysis above; and it gives us the denominator of our probability fractions.
(1) What is the probability of none of the passengers getting off on at least one bus stop?
Again this means the number of stops with no passengers getting off is 0 -- i.e., at least one passenger gets off at each stop.
To find the number of ways that can happen, we put one passenger at each stop; then we need to distribute the remaining 4 passengers among the 4 stops. Similar to above, the number of ways we can do that, using stars and bars, is
This also agrees with the number obtained earlier, so again the answer to this question is 35/165.
(2) And what is the probability of them not getting off on exactly 2 bus stops?
To find the number of ways of doing that, we need to do the following:
(a) choose 2 of the 4 stops to be the ones where no passengers get off
number of ways (using basic "n choose r"):
(b) place 1 passenger at each of the 2 stops where passengers do get off; then distribute the remaining 6 passengers among those two stops
number of ways (using stars and bars):
So the number of ways for the 8 passengers to all get off at exactly 2 of the stops (i.e., none get off at the other 2 stops) is 6*7=42.
And that agrees with the number we found with the earlier analysis, again giving us 42/165 as the answer to the second question.