Given a throw of three dice, let X,Y,Z be the number of dots showing on each. What is P(X<Y<Z)?
By symmetry this is
1/3! P(X,Y,Z are all different)
P(X,Y,Z are all different) =
1- P(X,Y,Z are not all different)
X,Y,Z are not all different if and only if one or more of the three conditions are satisfied:
c1: X = Y
c2: X = Z
c3: Y = Z
Terefore:
P(X,Y,Z are not all different)=
P(c1 OR c2 Or c3)
The prinicple of Inclusion and Exclusion says that:
P(c1 OR c2 Or c3) =
P(c1) + P(c2) + P(c3) -
[P(c1 And c2) + P(c1 And c3) +
P(c2 And c3) ] +
P(c1 And c2 And c3)
We have P(c1) = P(c2) = P(c3) = 1/6
and
P(c1 And c2) = P(c1 And c3) =
P(c2 And c3) = P(c1 And c2 And c3) = 1/36
Therefore:
P(c1 OR c2 Or c3) = 3/6 - 2/36 = 16/36 = 4/9
P(X,Y,Z are all different) = 1-4/9 = 5/9
P(X<Y<Z) = 1/6 * 5/9 = 5/54
Another way to compute this is by counting the number of ways you can make the three numbers diffrent. If you choose X first, then you have 6 choices for X. If Y is next then there are 5 choices for Y lft and 4 for Z. So, there are 6*5*4 possibilities. The probablity of any particular outcome is 1/6^3, so the probability is:
6*5*4/6^3 = 20/36 = 5/9
The probability that X, Y and Z are in the correct order is 1/6 times this probability which is:
5/9 * 1/6 = 5/54
To find the probability, P(X<Y<Z), we need to count the number of favorable outcomes and divide it by the total number of possible outcomes.
Let's break down the problem step by step:
First, let's consider the possible outcomes of throwing three dice. Each die has six faces, numbered 1 to 6. Therefore, the total number of possible outcomes is 6 * 6 * 6 = 216.
Now, let's determine the number of favorable outcomes where X<Y<Z.
To satisfy the condition X<Y<Z, we can't have any of the numbers equal. This means X must be less than Y and Y must be less than Z.
To count the favorable outcomes, we can start by considering the lowest possible value for X: 1. In this case, Y can take any value from 2 to 6, and Z can take any value from 3 to 6. So, the number of favorable outcomes when X=1 is (6-2) * (6-3) = 4 * 3 = 12.
Next, let's consider X=2. In this case, Y can take any value from 3 to 6, and Z can take any value from 4 to 6. So, the number of favorable outcomes when X=2 is (6-3) * (6-4) = 3 * 2 = 6.
Similarly, for X=3, the number of favorable outcomes is (6-4) * (6-5) = 2 * 1 = 2.
When X=4, there are no possibilities where X<Y<Z.
Finally, for X=5, there are no possibilities where X<Y<Z.
Adding up all the favorable outcomes: 12 + 6 + 2 + 0 + 0 = 20.
Therefore, the number of favorable outcomes where X<Y<Z is 20.
Finally, to find the probability, we divide the number of favorable outcomes by the total number of possible outcomes:
P(X<Y<Z) = 20/216 ≈ 0.0926
So, the probability of X being less than Y and Y being less than Z when throwing three dice is approximately 0.0926 or 9.26%.