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Question 2: Suppose that we roll pair of balanced dice and W is the number of dice that come up 3 or 6 and Z is the sum

Posted: Tue Aug 03, 2021 10:13 am
by answerhappygod
Question 2: Suppose that we roll pair of balanced dice and W is the number of dice that come up 3 or 6 and Z is the sum of the numbers on the dice. Find the joint probability distribution of Wand Z. Solution 2:


All outcome when we roll a pair of balenced dices are

(1,1),(1,2),(1,3),(1,4),(1,5) ,(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

Here W is the random variable for number of dice that comes up 3 or 6 .

W can be either 0 (when no 3 or 6 in two dices ) , 1 (when one of dice is 3 or 6) or 2 (when both dice are 3 or 6) .

Lets find outcome for each value of W

W= 0 ( no 3 or 6 in both dices)

(1,1),(1,2),(1,4),(1,5) ,(2,1),(2,2),(2,4),(2,5),(4,1),(4,2),(4,4),(4,5)(5,1),(5,2),(5,4), (5,5)

W= 1 (one of dice is 3 or 6)

(1,3) ,(1,6), (2,3), (2,6) ,(3,1),(3,2),(3,4),(3,5) ,(4,3),(4,6) , (5,3),(5,6) ,(6,1),(6,2),(6,4),(6,5)

W= 2 (both dice are 3 or 6)

(3,3) ,(3,6) ,(6,3), (6,6)

Here z is sum of number on dices .

It can be 2 ,3,4,5,6,7,8,9,10,11,12 .

So value of z are 2 ,3,4,5,6,7,8,9,10,11,12 .

Lets find outcome for each

z= 2 - (1,1)

z=3 -(1,2) ,(2,1)

z= 4 - (1,3),(3,1)(2,2)

z=5 - (1,4),(4,1),(2,3) ,(3,2)

z= 6 - (1,5) , (5,1) ,(2,4) ,(4,2) ,(3,3)

z= 7 - (1,6) ,(6,1),(2,5) ,(5,2) ,(3,4) ,(4,3)

z= 8 - (2,6),(6,2),(4,4),(3,5),(5,3)

z= 9 - (4,5),(5,4),(6,3),(3,6)

z=10 - (5,5),(6,4),(4,6)

z=11 - (5,6),(6,5)

z= 12 - (6,6)

Now we need to make joint probability distribution from this .

Lets find outcome for each combination and find probability

Z= 2 , W=0

Here (1,1) has sum 2 (Z=2) and has no 3 or 6 . So this is a outcome for this . Of 36 outcome only this gives z=2 , W=0 . So probability = 1/36

z= 2 , W= 1 , No outcome gives this . So probability =0

Z=2 , W= 2 . No outcome gives this . So probability =0

Z= 3 , W=0

Two outcome fro Z= 3 are (1,2),(2,1) . Both outcome has no 3 or 6 . So W= 0 .So probability= 2/36

z= 3 , W= 1 , No outcome gives this . So probability =0

Z=3, W= 2 . No outcome gives this . So probability =0

Z= 4 , W=0

The outcome (2,2) gives z= 4 and W=0 . So probability = 1/36

Z= 4 , W= 1

The outcome (1,3),(3,4) gives Z=4 and W=1 . So probability = 2/36

Z=4, W= 2 . No outcome gives this . So probability =0

Using same idea joint probability distribution table is as below

W
Z 0 1 2
2 36 0 0
3 2 36 0 0
4 36 2 36 0
5 2 36 2 36 0
6 4 36 0 36
7 2 36 4 36 0
8 36 4 36 0
9 2 36 0 2 36
10 36 2 36 0
11 0 2 36 0
12 0 0 36