Question

Let a fair die be rolled n times. Let’s make an assumption that all rolls of a die are independent. If x and y are the outcomes of any two of trials, what is the distribution function of xy?

Answer 1

\[nC2\times (\frac{1}{6})^2\times (\frac{5}{6})^{n-2}\]

Answer 2

I'm not sure I understand the way that you notated it

Answer 3

C is the choose function, are you familiar with it?

Answer 4

I know there are 36 outcomes hence (1/6)^2

Answer 5

not particularly, or not as notated there...

Answer 6

Ok, \(nCx=\frac{n!}{x!(n-x)!}\) is the amount of ways of choosing x things from n things without taking into account order. It's probably worth googling choose function/combinations.

Answer 7

right... I understand that... so in this particular instance I won't have multiple outcomes to show...the reason I say that is we had a prior question that was probability a is 2/3 find distribution function of it occurring in 9 trials

Answer 8

As I worked it out...
p9(0)=(2/3)^0*(1/3)^9=0
etc...

Answer 9

I guess my question is how would I notate it like that with two variables...?

Answer 10

And why "n-2" as the exponent after 5/6

Answer 11

Ok, I'm not sure I understand your question but I'll try to explain my answer nevertheless; you need the probability of an outcome which is 1/6 multiplied by 1/6 for the other outcome, this takes up two of your n trials so the remaining n-2 trials must all not be the outcome, you then multiply this by the number of ways of getting this outcome which is nC2. Sorry if that's not clear, look up 'Binomial Distribution' for more info.

Answer 12

I get what you're saying... 6 possible outcomes, if its xy, then that leaves the other outcomes (5/6) remaining and whatever n represents minus those two outcomes xy...

Answer 13

its the last question after have "real" numbers for everything else.... haha

Answer 14

Exactly!

Answer 15

Huh?

Answer 16

we werent doing distribution for "variables" on the rest of the hw...but rather set values of p and q...