Spot my mistake please...

In a class of 30 students,10 of them come to school by bus,12 students walk to school and the others by car. 6 students are randomly selected. Find the probability that

(a) only 1 student come to school by car (ans:0.3548)

(b) at least 2 students come to school by bus (ans: 0.6736)

(c) equal number of students from each category is selected. (ans: 0.1401)

My attempt:

(a) Probability that a student comes to school by car is 8/30=4/15

X~B(6,4/15)

P(X=1) = 6C1 x (4/15) x (11/15)^5

= 0.3393

(b) Probability that a student comes to school by bus is 10/30 = 1/3

Y~B(6,1/3)

P(Y=>2) = 1- 6C1 x (1/3) x (2/3)^5 - 6C0 x (1/3)^0 x (2/3)^6

= 0.6488

(c) Probability that a student comes to school on foot is 12/30=2/5

F~B(6,2/5)

P(2F,2Y,2X) = 6C2 x (2/5)^2 x (3/5)^4 x 4C2 x (1/5)^2 x (2/5)^2 x 1

= 0.01194

What is the mistake in my approach?