The question I need help understanding is below:

A balanced six-sided die is tossed twice. Let X be the highest number rolled.

a) Give the probability distribution for X.

The number of trials is two.

I need to figure out the probability that a certain number is the highest number displayed uppermost after a die is tossed, given that the die is tossed twice. So... If a six appears, what is the probability that it is the highest number of the two tossed? If I roll a six on one of the two tosses then it would seem more likely that it is the higher of the two numbers displayed. Is this thinking correct? How might I derive a binomial distribution from this problem?

Clearly the possible values of X are X = 1, 2, 3, 4, 5 and 6. So what rolls are favourable to each of these values and what's the probability of those rolls ....? Uing this dice table might help: Dice table

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I've never done a question where the number of trials is less than the value for X. From what I know, I'd have to take a negative factorial. If the die is tossed twice, does that mean that n=2?

I've found the probability of success and failure for each event X, and they're different for each value of X. Does this mean it is not considered a binomial experiment?

You need to think more about my reply. What rolls are favourable to X = 1? What's the probability of geting one of those rolls? What rolls are favourable to X = 2? What's the probability of geting one of those rolls? etc.

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The rolls favorable to X=1 include one for the first die, and one for the second die. The probability that one is the highest number displayed is 1/36.

There are three combinations for which two would be the highest number displayed for two dice rolls. X={1 2, 2 1, 2 2} The probability that two is the highest number displayed for such a roll is 3/36.

X=3, 5/36
X=4, 7/36
X=5, 9/36
X=6 , 11/36