00:01
So in this question we have a fair die which is rolled twice.
00:04
X is, so let's say that it's rolled twice, and we have a is, or a is the number on first roll, b is the number on the second roll, and a and b are both independently, independent and identically distributed according to the uniform distribution over the set 1, 2, 3, 4, 5 and 6.
00:41
So x is the absolute difference, a minus b, modulus, and y is a plus b.
00:51
So we want to find the joint pmf of x and y.
00:55
So the, so if, yeah, so let's try and work out what.
01:06
This is.
01:08
So if x is zero, so if x equals zero, then let's do this, let's do this in terms of conditional.
01:25
So let's do x, probability that x is equal to x and the probability that so y, so let's have x can be 0, well it can be 0, it can be 1, 2, 3, 4 or 5.
01:54
The probability that x equals that value is going to be.
02:00
So the probability x equals 0, they both have to be the same number.
02:05
So that's the probability a equals b.
02:07
And there's six ways of doing that over 36, so that's 1 .6th.
02:13
Probability x equals 1, there has to be a gap of 1 between them.
02:18
So we can think of pairs.
02:20
We've got 1, 2, 3, 4, 5 pairs, and each of those can be achieved in 2 ways, depending on which one we roll first.
02:29
So we've got 10 ways of doing it over 36.
02:34
For it to be 2, we have how many pairs now? 1, 2, 3, 4 pairs.
02:41
So there's 8 ways of doing it out of 36.
02:44
For a difference of three, we have one to three pairs, so six ways of doing it out of 36.
02:55
For a difference of four, there's only two pairs that can do the job.
03:02
So there's four ways of doing it out of 36.
03:05
And then for five, there's going to be two ways of doing it out of 36.
03:11
Okay.
03:11
So now if we have y, y can be 2, 3, 5, 6, 7, 8, 9.
03:34
Let's just summarise this a bit more simply.
03:41
So let's have our values of y across here.
03:46
And then the probability that y equals y given x equals x can go in the gap here.
03:51
So 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
04:00
So if the probability that y equals y given x equals 0, so for y to be equal to 2 when x equals 0, you need to have them both be 1.
04:10
So there's a 1 6th chance of that happening.
04:15
If x equals 0, then the two things are equal to each other, so y can't be odd.
04:23
For y to be equal to 4, they need to both be 2s, so these are all going to be 2.
04:27
To be one sixth.
04:33
For y to be two given x equals one, that can't happen.
04:37
So if x equals one, then the difference between the two things is one, and that means that the sum can't be even.
04:46
And we're going to find the same thing.
04:48
So when y is odd, x has to be odd, and when y is even, x has to be even.
04:59
So we'll get a situation like this.
05:15
So we don't have to fill in the whole table...