Probability Tables

A discrete random variable X takes a countable number of values (e.g., 0, 1, 2, 3, ...). A probability distribution table lists every possible value of X alongside its probability P(X = x).

For a valid probability distribution, two conditions must hold:

0 ≤ P(X = x) ≤ 1

Each probability is between 0 and 1

Σ P(X = x) = 1

All probabilities sum to 1

Example: Rolling a Fair Die

x	1	2	3	4	5	6
P(X = x)	1/6	1/6	1/6	1/6	1/6	1/6

Sum of probabilities: 6 × (1/6) = 1. Valid distribution.

Expected Value (Mean)

The expected value E(X) represents the long-run average value of a random variable. It is calculated as the weighted average of all possible values, using probabilities as weights.

Expected Value Formula

E(X) = Σ x · P(X = x)

For a fair die: E(X) = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 21/6 = 3.5. This means that over many rolls, the average result approaches 3.5.

Variance and Standard Deviation

Variance measures how spread out the values of X are from the mean. A higher variance means more spread.

Variance

Var(X) = E(X²) - [E(X)]²

Standard Deviation

SD(X) = √Var(X)

Where E(X²) = Σ x² · P(X = x). The standard deviation is in the same units as X, making it easier to interpret than variance.

Key Vocabulary

Discrete Random Variable

A variable that takes a countable number of distinct values (e.g., 0, 1, 2, ...).

Expected Value E(X)

The long-run average or mean of a random variable. E(X) = Σ x · P(X = x).

Variance Var(X)

A measure of spread around the mean. Var(X) = E(X²) - [E(X)]².

Probability Distribution

A table or function that assigns a probability to each possible value of a random variable.

Worked Examples

1

X has the distribution: P(0) = 0.3, P(1) = 0.5, P(2) = 0.2. Find E(X).

Step 1: E(X) = 0(0.3) + 1(0.5) + 2(0.2)

Step 2: = 0 + 0.5 + 0.4 = 0.9

Answer: E(X) = 0.9

2

Using the distribution from Example 1, find Var(X).

Step 1: E(X²) = 0²(0.3) + 1²(0.5) + 2²(0.2) = 0 + 0.5 + 0.8 = 1.3

Step 2: Var(X) = E(X²) - [E(X)]² = 1.3 - (0.9)² = 1.3 - 0.81 = 0.49

Answer: Var(X) = 0.49, SD(X) = √0.49 = 0.7

3

P(X = 1) = 0.2, P(X = 2) = 0.3, P(X = 3) = k. Find k if this is a valid distribution.

Step 1: All probabilities must sum to 1: 0.2 + 0.3 + k = 1

Step 2: k = 1 - 0.5 = 0.5

Answer: k = 0.5

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

X has: P(0) = 0.1, P(1) = 0.4, P(2) = 0.3, P(3) = 0.2. What is E(X)?

1.5 2.0 1.6 1.0

Question 2

For a valid probability distribution, what must all probabilities sum to?

0 0.5 It depends on the variable 1

Question 3

If E(X) = 4 and E(X²) = 20, what is Var(X)?

4 16 24 2

Question 4

P(X = 1) = 0.25, P(X = 2) = 0.35, P(X = 3) = k, P(X = 4) = 0.15. Find k.

0.15 0.25 0.30 0.75

Question 5

A game costs $2 to play. You win $5 with probability 0.3 and $0 otherwise. What is the expected profit?

$1.50 $0.50 -$0.50 -$1.50

Discrete Probability Distributions