$11$ kids get on a train with $3$ wagons. What's the probability that in the $1$st wagon there are exactly $3$ kids? Isn't this the same with saying $x_1+x_2+x_3=11$ and $x_1=3$? If yes, how could one solve this?
2026-04-21 20:17:41.1776802661
Kids in wagons probability
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1
Well, yes, it's the same, but neither question has enough information to answer it. You don't give the underlying probability distribution, so we can't evaluate the probability.
The "natural" choice of underlying distribution might be different depending on your two presentations of the question. In the $x_1 + x_2 + x_3 =11$ presentation, it might be natural to stipulate that all valid choices of $(x_1,x_2,x_3)$ are equally likely. In the children presentation, it might be natural to stipulate that for each child, getting onto any of the three trains is equally likely -- that's a different underlying distribution.
Assuming that that is the distribution in question, the children problem can be solved by noting that the probability of any specific setup in which 3 children are on the first train and 8 are on one of the other two is $\left(\frac13\right)^3\left(\frac23\right)^8$, and then multiplying this by the number of ways we can have 3 children on the first train, which is equivalent to the number of ways to choose 3 from 11, $\binom{11}3$.