Let "x" indicate the sum of the points on a die (which is nothing but the number of points on the die)
The sum/number of points on the dice would be
 1 if the dice shows up ONE
 2 if the dice shows up TWO
 3 if the dice shows up THREE
 4 if the dice shows up FOUR
 5 if the dice shows up FIVE
 6 if the dice shows up SIX
⇒ The values carried by the variable ("x") would be either 1, 2, 3, 4, 5 or 6
⇒ "X" is a discrete random variable with range = {1, 2, 3, 4, 5, 6}
"X" represents the random variable and P(X = x) represents the probability that the value within the range of the random variable is a specified value of "x"
In the experiment of tossing a dice, there are six possible elementary events, the events of the die showing up either ONE, TWO, THREE, FOUR, FIVE or SIX all of which are mutually exclusive, equally likely and exhaustive.
Therefore the probability of occurance of each elementary event is 1/6
Probabilty that the dice would show up

ONE ⇒ P(ONE) 
= 


TWO ⇒ P(TWO) 
= 


THREE ⇒ P(THREE) 
= 


FOUR ⇒ P(FOUR) 
= 


FIVE ⇒ P(FIVE) 
= 


SIX ⇒ P(SIX) 
= 

Probability for the sum/number of points on the dice to be

1 ⇒ P(X = 1) 
= 


2 ⇒ P(X = 2) 
= 


3 ⇒ P(X = 3) 
= 


4 ⇒ P(X = 4) 
= 


5 ⇒ P(X = 5) 
= 


6 ⇒ P(X = 6) 
= 

The probabilty distribution of "x" would be
Calculations for Mean and Variance

x 
P (X = x) 
px [x × P (X = x)] 
x^{2} 
px^{2} [x^{2} × P (X = x)] 

1 


1 


2 


4 


3 


9 


4 


16 


5 


25 


6 


36 

Total 

1 






= 3.5 

= + 15.17 
Expectation of the sum/number of points on the dice
⇒ Expectation of "x"
Variance of the sum/number of points on the die
⇒ var (x) 
= 
E (x^{2}) − (E(x))^{2} 
⇒ var (x) 
= 
Σ px^{2} − (Σ px)^{2} 

= 
15.17 − (3.5)^{2} 

= 
15.17 − 12.25 

= 
2.92 
Let x_{1}, x_{2} represent the sum of the points on the first die and the second dice respectivley
Expected sum/number of the points
 On the First Dice
⇒ E (x_{1}) = 3.5
 On the Second Dice
⇒ E (x_{2}) = 3.5
Each trial (throwing of the dice) is identical and therefore the expected sum/number of points on the dice in each trial would be the same
Therefore, expected sum of the points on the two dice
= E (x_{1}) + E (x_{2})
= 3.5 + 3.5
= 2 × 3.5
= 7
Variance of the sum/number of points
 On the First Dice
⇒ var (x_{1}) = 2.92
 On the Second Dice
⇒ var (x_{2}) = 2.92
Each trial (throwing of the dice) is identical and therefore the variance of the sum/number of points on the dice in each trial would be the same
Variance of the sum of the points on the two dice
= var (x_{1}) + var (x_{2})
= 2.92 + 2.92
= 2 × 2.92
= 5.84
Alternative
Where all the trials are identical
 The expected sum of the points is given by
Number of Trials × Expectation in each trial
⇒ Expected sum of the points on two dice
= No. of dice thrown × Expected sum of the points per dice
= n × E (x)
= 2 × 3.5
= 7

The variance of the sum of the points is given by
Number of Trials × variance in each trial
⇒ variance of the sum of the points on two dice
= No. of dice thrown × variance of the sum of the points on the two dice
= n × var (x)
= 2 × 2.92
= 5.84
Alternative
Let "x" indicate the sum of the points on the two dice
[Since you are required to find the expected value the sum of the scores on the two dice, the variable would represent the sum of the points on the dice]
The sum of the points on the two dice would be
 2 if the dice show up {(1,1)}
 3 if the dice show up {(1,2), (2,1)}
 4 if the dice show up {(1,3), (2,2), (3,1)}
 5 if the dice show up {(1,4), (2,3), (3,2), (4,1)}
 6 if the dice show up {(1,5), (2,4), (3,3), (4,2), (5,1)}
 7 if the dice show up {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
 8 if the dice show up {(2,6), (3,5), (4,4), (5,3), (6,2)}
 9 if the dice show up {(3,6), (4,5), (5,4), (6,3)}
 10 if the dice show up {(4,6), (5,5), (6,4)}
 11 if the dice show up {(5,6), (6,5)}
 12 if the dice show up {(6,6)}
⇒ The values carried by the variable ("x") would be either 1, 2, 3, ... , 12
⇒ "X" is a discrete random variable with range = {1, 2, 3, ... , 12}
"X" represents the random variable and P(X = x) represents the probability that the value within the range of the random variable is a specified value of "x"
In the experiment of tossing 2 dice
Total no. of possible choices 
= 
6 × 6 

= 
36 
Probabilty that the two dice would show up

{(1,1)} 
= 
No. of favorable choices  Total no. of possible choices 


{(1,2), (2,1)} 
= 
No. of favorable choices  Total no. of possible choices 


{(1,3), (2,2), (3,1)} 
= 
No. of favorable choices  Total no. of possible choices 


{(1,4), (2,3), (3,2), (4,1)} 
= 
No. of favorable choices  Total no. of possible choices 


{(1,5), (2,4), (3,3), (4,2), (5,1)} 
= 
No. of favorable choices  Total no. of possible choices 


{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} 
= 
No. of favorable choices  Total no. of possible choices 


{(2,6), (3,5), (4,4), (5,3), (6,2)} 
= 
No. of favorable choices  Total no. of possible choices 


{(3,6), (4,5), (5,4), (6,3)} 
= 
No. of favorable choices  Total no. of possible choices 


{(4,6), (5,5), (6,4)} 
= 
No. of favorable choices  Total no. of possible choices 


{(5,6), (6,5)} 
= 
No. of favorable choices  Total no. of possible choices 


{(6,6)} 
= 
No. of favorable choices  Total no. of possible choices 

Probability that the sum of the numbers on the two dice would be

2 ⇒ P(X = 2) 
= 


3 ⇒ P(X = 3) 
= 


4 ⇒ P(X = 4) 
= 


5 ⇒ P(X = 5) 
= 


6 ⇒ P(X = 6) 
= 


7 ⇒ P(X = 7) 
= 


8 ⇒ P(X = 8) 
= 


9 ⇒ P(X = 9) 
= 


10 ⇒ P(X = 10) 
= 


11 ⇒ P(X = 11) 
= 


12 ⇒ P(X = 12) 
= 

The probabilty distribution of "x" would be
x 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
P(X = x) 











Calculations for Mean and Variance

x 
P (X = x) 
px [x × P (X = x)] 
x^{2} 
px^{2} [x^{2} × P (X = x)] 

2 


4 


3 


9 


4 


16 


5 


25 


6 


36 


7 


49 


8 


64 


9 


81 


10 


100 


11 


121 


12 


144 

Total 

1 






= 7 

= + 54.83 
Expectation of the sum of the points on the dice
⇒ Expectation of "x"
Variance of the sum of the points on the two dice
⇒ var (x) 
= 
E (x^{2}) − (E(x))^{2} 
⇒ var (x) 
= 
Σ px^{2} − (Σ px)^{2} 

= 
54.83 − (7)^{2} 

= 
54.83 − 49 

= 
5.83 
Standard Deviation of the sum of the points on the dice
⇒ SD (x) 
= 
+ √ Var (x) 

= 
+ √ 5.83 

= 
+ 2.412 
