As @Andre Nicolas suggested, I used principle of inclusion and exclusion.
Total number of ways $=3^{10}$
But I have also counted the number of ways where one bus remains empty. I need to subtract those cases. Number of ways where one bus remains empty $=\binom{3}{1}2^{10}$, where $\binom{3}{1}$ is for choosing the bus.
But I have subtracted the case where 2 buses are empty twice. I need to add that back. Number of ways where 2 buses are empty $=\binom{3}{2}1^{10}$
$$\text{probability}=\frac{3^{10}-\binom{3}{1}2^{10}+\binom{3}{2}1^{10}}{3^{10}}$$
The mistake I made was pointed out by @Andre Nicolas in a comment to the original post:
Unfortunately your method involves double-counting or worse. The flaw is with the idea of putting certain passengers on "first." You have treated Alicia being "first" on bus X, and Beti also being a passenger, as different from Beti being put first on bus X, and Alicia also being a passenger. A correct method is Inclusion/Exclusion. – André Nicolas
Let's ask two different questions:
How many ways can 150 people be distributed among 5 train cars such that no more than 50 are on any given car?
If there were infinite seats on each train car, how many ways could 150 people be distributed amongst 5 train cars?
If each person is choosing independently of the other, then the answer would be the ratio of the first question over the second question.
The second problem is easy: It's 5^150.
The first problem is more interesting. Think partitions of n into k parts each less than or equal to d. However, we can get an easier sums by remembering the following facts:
The number of people in three cars must be at least 50 and at most 150.
The number of people in two cars can be any number from 0 to 100.
We can treat the people as indistinguishable if we multiply the final result by the permutations of the people ($50!$).
Using these, we can split the train cars up into groups to get an easy sum.
Let's split into two cases: First, if there are no more than 50 people in the first two cars, we can write this as $$\sum_{x=a+b=0}^{50}(x+1)\sum_{c=50-x}^{50}(c+x-49)$$
In other words, we choose x people, up to 50 are in the first two cars, and there are $x+1$ ways to divide them between the two cars (from 0 in the first and x in the second to x in the first and 0 in the second). Then, with the remaining 150-x people, we pick a number for the third car $c$, which must be at least $50-x$. Once we've done that, since we have at most 100 people in the first three cars, we have at least 50 people to distribute between the last two. If there are 100 people, we have 1 way to distribute and if we have 50, we have 51 ways. This works out to $101-p$ ways to distribute $p$ people, which simplifies to $c+x-49$.
Second case: There are more than 50 people in the first two cars. Then we don't have to limit our choices on the last three cars, only on the second one. If there are $a=1$ to $50$ people in the first car, there must be at least $b=51-a$ people in the second car. We can pick any number for the third car. Then there are $150-c-b-a$ people to distribute between the last two cars, and we'll have to split our cases again.
Let $y=150-a-b-c$, the number of people we're putting into the last two cars. If $100\geq y>50$, then the number of ways to split the folks is $101-y$. If $50\leq y$, the number of ways to split the folks is $y+1$. So this gives us the following sum:
$$\sum_{a=1}^{50}\sum_{b=51-a}^{50}\left[\sum_{c=0}^{100-a-b} (a+b+c-49)+ \sum_{c=101-a-b}^{50} (151-a-b-c)\right]$$
Using summation tricks, both of these are evaluatable. Maple gives me a total of 3134001 for the two sums. Multiplying by $50!$ and dividing by the second question gives the probability: $1.36*10^{(-34)}$.
Best Answer
What you have computed as $P_2$ is actually $N_2$, the # of ways.
However, it won't matter if we are just to compute the ratio $\dfrac{P_1}{P_2}$
Making the same assumption as you, that the six cars are distinct,
$\dfrac{P_1}{P_2} = \dfrac{\dbinom{12}{0,1,2,2,3,4}}{\dbinom{12}{2,2,2,2,2,2,}}= \dfrac{(2!)^4}{3!4!}$
PS.
Since the cars are taken to be distinct, the numerator needs to be multiplied by $\dfrac{6!}{2!}$ to take care of permutations of the pattern of distribution. You will then get the answer of $40$.