In: Statistics and Probability
3. Let the experiment be the toss of three dice in a row. Let X be the outcome of the first die. Let Y be the outcome of the 2nd die. Let Z be the outcome of the 3rd die. Let A be the event that X > Y , let B be the event that Y > Z, let C be the event that Z > X.
(a) Find P(A).
(b) Find P(B).
(c) Find P(A ∩ B).
(d) Are A and B independent?
(e) Are A, B, C pairwise independent?
(f) Find P(A ∩ B ∩ C).
(g) Are A, B, C mutually independent?
In this experiment the total number of outcomes are .
(a) The event A says X>Y but it doesn't put any conditions on Z. So all values of Z are considered.
So for X=x there will be (x-1) possible values of Y in the set {6,5,4,3,2,1}, where x=2,3,4,5,6.
And for every chosen pair (x,y) there are 6 possible values of Z.
So the number of favourable outcomes is
(b) The problem is similar as (a) just instead of X>Y, we have Y>Z and here X can take any of 6 possible values.
So the required probability is same as above, i.e. .
(c) Now
So for X=x, Y can be chosen from the set {2,3,...,(x-1)} and for Y=y there will be (y-1) possible values of Y in the set {6,5,4,3,2,1}, where x=3,4,5,6 and y=2,3,...,(x-1).
So the number of favourable outcomes is
(d) Since , A and B are not independent.
(e) Since the cases for and are similar to the case of , the results will be same.
So B and C are not independent and A and C are not independent.
So A,B,C are not pairwise independent.
(f) Now,
So this is an impossible event Since X>Z and Z>X can't happen at same time.
(g)
Since A,B,C are not mutually independent.