What is the probability that less than four of the balls are numbered 2 I'm a high school student and I'm struggling with one problem we've been given. Since we are allowed to use calculators, TI84's I am kinda confused If I can use in this problem binomcdf or is there any other way how can I calculate it?

Six balls numbered 1,2,2,3,3,3 are placed in a bag. Balls are taken one at a time from the bag at random and the number noted. Throughout the question a ball is always replaced before the next ball is taken. 

-- The question is: Ten balls are taken from the bag. Find the probability that less than four of the balls are numbered 2.
the probability is (the number of events that you want/the number of all events):
the number of all events is 6^10
events that you want:
1)3 balls are numbered 2:
-Suppose that the first 3 ball are numbered 2 and The another 7 ball would not be numbered 2, like this: 2,2,2,X,X,X,X,X,X,X. For each X there are 4 possibilities, then in this case there would have $4^{7}$ possibilities. There are $C_{10,3}$
ways to choose the positions of balls numbered 2, each of way with $4^{7}$ possibilities, so the total of cases with 3 ball  numbered 2 is $4^{7}*C_{10,3}$

2)2 balls are numbered 2:
In a similar way: $4^{8}*C_{10,2}$

1)1 ball are numbered 2:
In a similar way: $4^{9}*C_{10,1}$

4)0 ball are numbered 2:
$4^{10}$

the result is $\frac{4^{7}*C_{10,3}+4^{8}*C_{10,2}+4^{9}*C_{10,1}+4^{10}}{6^{10}}$
the probability is (the number of events that you want/the number of all events):
the number of all events is 6^10
events that you want:
1)3 balls are numbered 2:
-Suppose that the first 3 ball are numbered 2 and The another 7 ball would not be numbered 2, like this: 2,2,2,X,X,X,X,X,X,X. For each X there are 4 possibilities, then in this case there would have $4^{7}$ possibilities. There are $C_{10,3}$
ways to choose the positions of balls numbered 2, each of way with $4^{7}$ possibilities, so the total of cases with 3 ball  numbered 2 is $4^{7}*C_{10,3}$

2)2 balls are numbered 2:
In a similar way: $4^{8}*C_{10,2}$

1)1 ball are numbered 2:
In a similar way: $4^{9}*C_{10,1}$

4)0 ball are numbered 2:
$4^{10}$

the result is $\frac{4^{7}*C_{10,3}+4^{8}*C_{10,2}+4^{9}*C_{10,1}+4^{10}}{6^{10}}$