00:01
In this problem, we're rolling three six -sided dye, and part a is asking us to find the probability that we get triples.
00:08
So triples means it could be three -ones or three -twos or three -threes.
00:13
So suppose we have three -ones, that would be one -six times one -six times one -six, or it could be three - twos, in which case it would be the same thing, one -six, one -six, or three -threes and so on.
00:29
So actually we're going to get six different options here.
00:35
The ones, two, three, fours, fives, and sixes.
00:38
So this will end up being one over 36.
00:43
In part b, this is a conditional probability.
00:48
So first of all, we are only going to consider all the options that have a sum less than are equal to eight, and we're going to have to generate that in a second.
00:57
The numerator is considering the number of triples.
01:00
So think about the number of triples that would be less than or equal to 8.
01:04
Well, 1 -1 -21 -22.
01:07
333 is already a sum of 9.
01:09
So there's only two options of triples for us to consider, but we've got to figure how many options we're considering from.
01:17
So we actually have to map this out, and we're going to do this on the bottom right here.
01:23
Let's list sample space.
01:25
Some less than or equal to 8.
01:26
Well, we could have 111, 112, 113, 114, 115, 116.
01:36
And i'm going to recognize here that these guys all have three different arrangements.
01:45
Could be, let's take 116, for example, that could be 161 or 611.
01:51
Those are all different combinations, or different permutations.
01:54
So we would have three times this many number of combinations with these sets of numbers.
02:02
But in addition to that, we could also have starting with a 1 and then 2 -2.
02:10
1 -2 -3, 1 -2 -4.
02:14
That would add up to 7.
02:16
1 -2 -5.
02:19
And let's try 1 -3 -13 -1 -3.
02:23
1 -3 -4 is okay as well.
02:27
Because that would add up to eight and we can do one four one which we've already counted because it's one one four one four two is counted one four two is counted one four three we've done already and one four would exceed that so this is it that's the all the combinations that start with you know that have a one in them so this one will have three combinations these ones are three distinct objects so there will be six combinations is it three factorial.
03:00
This will have three combinations and three distinct objects here will have six combinations.
03:07
Now let's start with two.
03:09
Two, two, two, two, two, two, two, three, two, four.
03:16
We can't go to two, five, because that would exceed.
03:19
We could have two, three, one, which if we've done already, 232, which is the same as 223.
03:35
So 233, we have not done.
03:42
And we can't go to 234.
03:44
We can go 241, which we've done.
03:49
242 is done.
03:52
243 would already exceed.
03:54
So this is it.
03:56
So 223, 224...