A person picking 3 balls from a bag can win Rs. 10 for each blue ball drawn and lose Rs. 5 for each red ball drawn.
In the experiment of drawing 3 balls from a bag containing 5 red and 4 blue balls, where "R" represents a red ball and "B" represents a blue ball
The sample space S = (BBB, BBR, BRB, RBB, RRB, RBR, BRR, RRR}
If "s' represents the elements of the set "S" and "x" the number of heads obtained
Then, the random variable "X" representing the relationship between "s" and "x" is
X(s) = x (Or) X : s → x
The value of "x" as determined by the outcome of the experiment would be
Event [Balls drawn being] 
BBB 
RBB 
BRB 
BBR 
RRB 
RBR 
RRB 
RRR 
"x" 
+ 30 
+ 15 
+ 15 
+ 15 
0 
0 
0 
− 15 
"X" is a function with Domain "S" and range = {−15, 0, +15 or +30}
The number of elements in the sample space is 8 (i.e. n(s) = 8) and all the events are all equally likely and mutually exclusive,
⇒ Probability of occurrence of each elementary event is 1/n i.e. 1/8
The probabilities of various events would be
Event [Coins Showing Up] 
HHH 
THH 
HTH 
HHT 
TTH 
THT 
TTH 
TTT 
P(Event)" 








The probabilities of occurrence of the different values in the range of the random variable would be
P(X = +30) 
= 
P (BBB) 

⇒ P(X = +30) 
= 

P(X = +15) 
= 
P(BBR) + P(BRB) + P(RBB) 
= 

⇒ P(X = +15) 
= 

P(X = 0) 
= 
P(BRR) + P(RBR) + P(RRB) 
= 

⇒ P(X = 0) 
= 

P(X = −15) 
= 
P (RRR) 

⇒ P(X = −15) 
= 

P(X = 0) = P (BRR) + P(RBR) + P(RRB)
Probability of getting a blue ball and 2 red balls
⇒ Probability that at least one of these events should occur
⇒ P (BRR U RBR U RRB) = P (BRR) + P(RBR) + P(RRB)
["BRR", "RBR", "RRB" are elementary events which are mutually exclusive, ]
The distribution indicating the numerical values in the range of "X" and their respective probabilities of occurrence
This distribution indicating all the possible numerical values (within the range of a discrete random variable in relation to an experiment) with their respective probabilities is called a discrete probability distribution.
