Question

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?

Answer 1

In this experiment the total number of outcomes are   $\small 6^3=216$ .

(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

6 6 1) =-Σε-Σ6= 6 90 6(x x 20) 30 =2 -2 r-2

5 P(A) 90/216 12

(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.   .

P(B) 12

(c) Now  

(An B) = (X > Y,Y > Z) = (X> Y> Z)

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

x-1 x-1 6 -1 6 6 Σι ΣΣυ-)-ΣΙΣ 1 ν α=3 y=2 3 y-2 α=3 y=2 ΑΤ

6 6 (π-2)(2+a-1) 1 ΣΣ-Σ: - Σ-2) ο-2 Σ Σ 2 2 _ 2 -3 r-3 --μ

0.5(86 18 8) (18 8) 20

20 5 . P(An B) 54 216

(d) Since   ,   A and B are not independent.

PAn B)PA P(B)

(e) Since the cases for  
(BnC)
   and  
Cn A) Y
   are similar to the case of   $\small (A\cap B)$ , 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,

(An BnC) (X> Y, Y> Z, Z> X) = (X> Y> Z, Z> X) =

So this is an impossible event Since X>Z and Z>X can't happen at same time.

$\small \therefore P(A\cap B\cap C)=0$

(g)   $\small P(C)=P(B)=P(A)=\frac{5}{12}$

Since   $\small P(A\cap B\cap C) \neq P(A)P(B)P(C)$    A,B,C are not mutually independent.