AS maths binomial distribution question

A fair six-sided die is rolled six times. Which of the following is/are given by 6C2(1/6)^2(5/6)^4 ?
A: The probability of getting exactly one 2.
B: The probability of getting exactly two 1s.
C: The probability of getting exactly four 5s.
D: The probability of getting exactly five 4s.
E: The probability of getting exactly two scores less than 6.
F: The probability of getting exactly four scores greater than 1

A fair six-sided die is rolled six times. Which of the following is/are given by 6C2(1/6)^2(5/6)^4 ?
A: The probability of getting exactly one 2.
B: The probability of getting exactly two 1s.
C: The probability of getting exactly four 5s.
D: The probability of getting exactly five 4s.
E: The probability of getting exactly two scores less than 6.
F: The probability of getting exactly four scores greater than 1

Its obviously from a binomial distribution, so any ideas?

I can see that B would be 6C2(1/6)^2(5/6)^4, but I'm confused about E and F

Agree about that. For the last two, how would you write down the probabilities?

For E, would it be 6C2(5/6)^2(1/6)^4
And for F would it be 6C4(5/6)^4(1/6)^2

Agree about that. For the last two, how would you write down the probabilities? Are they equivalent to any of A-D?

And you know 6C2 = 6C4?
F must be the same as B, even if you don't understand probabilities?

I don't think the last two would be equivalent to any of A-D, because the success probability is 1/6 for A to D, but this changes to 5/6 for E and F

Can you see which one E is equivalent to? It's not identical, but essentially the same.

Yes, is the probability of E equivalent to C?

Yes. When using a binomial, you can go down the p or 1-p route, and note the nCr symmetry
nCr = nC(n-r)
Which is the symmetry in Pascal triangle.