# Discrete Probability Distribution of a Random (Stochastic or Chance) Variable

### Probability Distribution

Distribution = An arrangement of values of a variable showing their frequency of occurrence

Each elementary event in a random experiment (sample point in the sample space) has a certain probability of occurrence. The random variable associates each event of the experiment with a numerical value. The set of possible numerical values related to an experiment forms the range of the random variable.

Therefore, each numerical value in the range has a certain probability of occurrence associated with it.
Eg: 1. In the experiment of tossing 3 coins, where "H" represents a head on a coin and "T" a tail on a coin,

The sample space S = (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

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 [Coins Showing Up]       "x" HHH THH HTH HHT TTH THT TTH TTT 3 2 2 2 1 1 1 0

"X" is a function with domain "S" and Range = {0, 1, 2, 3}.

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)
 1 8
 1 8
 1 8
 1 8
 1 8
 1 8
 1 8
 1 8

The probabilities of occurrence of the different values in the range of the random variable would be
P(X = 0) = P (TTT) ⇒ P(X = 0) =
 1 8
P(X = 1) = P (HTT) + P(THT) + P(TTH) ⇒ P(X = 1) =
 1 8
+
 1 8
+
 1 8
⇒ P(X = 1) =
 3 8
P(X = 2) = P (HHT) + P(HTH) + P(THH) ⇒ P(X = 2) =
 1 8
+
 1 8
+
 1 8
⇒ P(X = 2) =
 3 8
P(X = 3) = P (HHH) ⇒ P(X = 3) =
 1 8

P(X = 2) = P (HHT) + P(HTH) + P(THH)

⇒ Probability that at least one of these events should occur

⇒ P (HHT U HTH U THH) = P (HHT) + P(HTH) + P(THH)

["HHT", "HTH", "THH" are elementary events which are mutually exclusive,]

The distribution indicating the numerical values in the range of "X" and their respective probabilities of occurrence
x 0 1 2 3
P(X=x)
 1 8
 3 8
 3 8
 1 8

This distribution indicating all the possible numerical values (within the range of a random variable in relation to an experiment) with their respective probabilities is called a probability distribution.

### Discrete Probability Distribution

The distribution indicating the possible numerical values (within the range of a discrete random variable related to an experiment) and their respective probabilities is called a "Discrete Probability Distribution"

If "X" is a discrete random variable with range x1, x2, x3, ...., xn with respective probabilities p1, p2, p3, ...., pn, where p1 + p2 + p3 + ....+ pn = 1, then the set P(X=xi) = p(xi) is called the probability distribution of the discrete random variable "X". It is generally represented as a table with the range of values of the discrete random variable and their respective probabilities together.
Eg: 1. In the experiment of rolling a die,

The sample space S = (ONE, TWO, THREE, FOUR, FIVE, SIX}

The number of elements in the sample space is 6 (i.e. n(s) = 6) and all the events are all equally likely and mutually exclusive,

⇒ Probability of occurrence of each elementary event is 1/n i.e. 1/6

⇒ P(ONE) = P(TWO) = P(THREE) = P(FOUR) = P(FIVE) = P(SIX) = 1/6

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

This relationship can be summarised as
 Event [Die Showing Up]       "x" ONE TWO THREE FOUR FIVE SIX 1 2 3 4 5 6

"X" is a function with domain "S" and Range = {1, 2, 3, 4, 5, 6}.

The probabilities of occurrence of the different values in the range of the random variable would be
P(X = 1) = P (ONE) ⇒ P(X = 1) =
 1 6
P(X = 2) = P (TWO) ⇒ P(X = 2) =
 1 6
P(X = 3) = P (THREE) ⇒ P(X = 3) =
 1 6
P(X = 4) = P (FOUR) ⇒ P(X = 4) =
 1 6
P(X = 5) = P (FIVE) ⇒ P(X = 5) =
 1 6
P(X = 6) = P (SIX) ⇒ P(X = 6) =
 1 6

The distribution indicating the numerical values in the range of "X" and their respective probabilities of occurrence
x 1 2 3 4 5 6
P(X=x)
 1 6
 1 6
 1 6
 1 6
 1 6
 1 6

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.

2.   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]       "x" BBB RBB BRB BBR RRB RBR RRB RRR + 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)"
 1 8
 1 8
 1 8
 1 8
 1 8
 1 8
 1 8
 1 8

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) =
 1 8
P(X = +15) = P(BBR) + P(BRB) + P(RBB) =
 1 8
+
 1 8
+
 1 8
⇒ P(X = +15) =
 3 8
P(X = 0) = P(BRR) + P(RBR) + P(RRB) =
 1 8
+
 1 8
+
 1 8
⇒ P(X = 0) =
 3 8
P(X = −15) = P (RRR) ⇒ P(X = −15) =
 1 8

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
x − 15 0 +15 +30
P(X=x)
 1 8
 3 8
 3 8
 1 8

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.

 Author Credit : The Edifier ... Continued Page 4