No the number that come by car does not have a binomial distribtion.

The probability that the first student came by car is 8/30, then that the next did not come by car is 22/29, and then next 21/28, ...

So the required probbailty that the first came by car and the rest did not is:

(8x22x21xx20x19x18)/(30x29x29x27x26x27).

The total probabilty you require is six times this (to account for the cases where the child coming by car is the 1st, 2nd, ..., 6th).

CB