Probability Distribution

Possible Values and Their Probabilities

A probability distribution lists the possible values of a random quantity and assigns a probability to each value. For a die, the quantity might be the number shown. For two dice, it might instead be their sum. The distribution tells us which values are possible and how likely each one is.

The Sample Space of All Possible Outcomes

The sample space is the set of every elementary outcome that can occur in an experiment. We write it with the symbol $\Omega$ (Omega).

For one roll of a standard $6$ -sided die, the possible outcomes are $1, 2, 3, 4, 5,$ and $6$ . Together they form the sample space:

\Omega = \{1, 2, 3, 4, 5, 6\}

There are a total of $6$ possible outcomes here.

Events and Their Probabilities

Event

An event is a subset of the sample space containing the outcomes that satisfy a condition of interest.

Example:

For the die experiment ( $\Omega = \{1, 2, 3, 4, 5, 6\}$ ), consider the event "the result is even." The even outcomes are $2, 4,$ and $6$ , so event $A$ is:

A = \{2, 4, 6\}

There are $3$ outcomes in this event $A$ .

Calculating the Probability of an Event

Probability measures how likely an event is to occur.

When all elementary outcomes are equally likely, the probability is the number of favorable outcomes divided by the total number of outcomes:

P(\text{Event}) = \frac{\text{Number of outcomes in that event}}{\text{Total number of outcomes in the sample space}}

Using the symbols from our example:

P(A) = \frac{|A|}{|\Omega|}

Here:

$|A|$ is the number of outcomes in event $A$ , which is $3$ .
$|\Omega|$ is the total number of outcomes in the sample space, which is $6$ .

So, the probability of the event "getting an even number" (event $A$ ) is:

P(A) = \frac{3}{6} = \frac{1}{2}

The event therefore has probability $1/2$ , or $50\%$ .

Rules for Probability Distributions

A valid probability distribution satisfies two rules:

The probability of each value ( $P(x)$ ) lies between $0$ and $1$ .

$0 \leq P(x) \leq 1$
The probabilities of all possible values add up to exactly $1$ .

$\sum_{x \in \Omega} P(x) = 1$

Rolling One Die

For one fair die, every face has the same probability. Its distribution is:

Outcome $x$	Probability $P(x)$
$1$	$1/6$
$2$	$1/6$
$3$	$1/6$
$4$	$1/6$
$5$	$1/6$
$6$	$1/6$

Why are they all $1/6$ ?

The die has $6$ equally likely faces, so each face has probability $1/6$ . Their sum is $1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 = 1$ , as required.

Rolling Two Dice

Now roll one red die and one white die. Listing the result as an ordered pair produces $36$ possible outcomes.

Why $36$ ?

The red die has $6$ possibilities and the white die has $6$ possibilities, giving $6 \times 6 = 36$ ordered pairs.

The resulting pairs can be written like this:

Red die $1$ , white die $1$ gives $(1, 1)$ .
Red die $1$ , white die $2$ gives $(1, 2)$ .
... and so on until ...
Red die $6$ , white die $6$ gives $(6, 6)$ .

Because the dice are fair and independent, each ordered pair has probability $1/36$ .

Distinguish $(1, 2)$ from $(2, 1)$

Because the dice have different colors, a red $3$ with a white $2$ differs from a red $2$ with a white $3$ . The order records which die produced each number.

Probability Distribution for the Sum of Two Dice

If the random variable is the sum of the two dice, its smallest value is $1+1=2$ and its largest value is $6+6=12$ . Different sums contain different numbers of ordered pairs, so their probabilities differ:

Sum of Numbers $j$	Possible Pairs	Number of Pairs	Probability $P(j)$
$2$	$(1,1)$	$1$	$1/36$
$3$	$(1,2), (2,1)$	$2$
$4$	$(1,3), (2,2), (3,1)$	$3$
$5$	$(1,4), (2,3), (3,2), (4,1)$
$6$	$(1,5), (2,4), (3,3), (4,2), (5,1)$
$7$	$(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$
$8$	$(2,6), (3,5), (4,4), (5,3), (6,2)$
$9$	$(3,6), (4,5), (5,4), (6,3)$
$10$	$(4,6), (5,5), (6,4)$	$3$
$11$	$(5,6), (6,5)$	$2$
$12$	$(6,6)$	$1$	$1/36$

The sum $7$ is most likely because $6$ ordered pairs produce it, while the sums $2$ and $12$ each come from only one pair. This difference matters when a game or model uses the total from both dice.

Comparing Outcomes and Risks

A probability distribution pairs each possible outcome with its probability. Those numbers support several concrete comparisons:

Compare outcomes: A sum of $7$ has a higher probability than a sum of $12$ .
Analyze games: In games that use dice, the distribution shows which moves or totals occur more often, even though the result of one roll remains uncertain.
Make predictions: A distribution can summarize uncertainty, such as the probability of rain under a weather model.
Support decisions: Science and business use probability distributions to compare risks and expected outcomes with evidence from data.

Each use relies on the same structure: a defined set of outcomes and a probability assigned to every outcome.

Exercise

Three fair coins are tossed. Let $X$ be the number of heads. Build the probability distribution of $X$ , verify that it is valid, and find $P(X \geq 2)$ .

Worked Solution

Encode a head as $1$ and a tail as $0$ . The three tosses then produce $2^3=8$ equally likely ordered triples:

\Omega=\{(1,1,1),(1,1,0),(1,0,1),(0,1,1),(1,0,0),(0,1,0),(0,0,1),(0,0,0)\}

Count the outcomes that contain $0,1,2$ or $3$ heads:

Number of heads $x$	Outcomes	Count	$P(X=x)$
$0$	$(0,0,0)$	$1$	$1/8$
$1$	$(1,0,0),(0,1,0),(0,0,1)$
$2$	$(1,1,0),(1,0,1),(0,1,1)$
$3$	$(1,1,1)$	$1$	$1/8$

Every probability lies between $0$ and $1$ , and their sum is

\frac{1}{8}+\frac{3}{8}+\frac{3}{8}+\frac{1}{8}=\frac{8}{8}=1

The distribution is therefore valid. At least two heads means $X=2$ or $X=3$ , so

P(X \geq 2)=P(X=2)+P(X=3)=\frac{3}{8}+\frac{1}{8}=\frac{1}{2}

Listing the sample space first makes the distribution auditable: each of the $8$ outcomes appears in exactly one row.