IB Mathematics AA Discrete and continuous random variables and their probability distributions Study Notes
IB Mathematics AA Discrete and continuous random variables and their probability distributions Study Notes
IB Mathematics AA Discrete and continuous random variables and their probability distributions Notes Offer a clear explanation of Discrete and continuous random variables and their probability distributions, including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Discrete and continuous random variables and their probability distributions.
Concept of Discrete Random Variables and Their Probability Distributions
Concept of Discrete Random Variables and Their Probability Distributions
A discrete random variable is a variable that takes specific, separate (discrete) values, each associated with a probability.
For example, when rolling a die, the number showing on the top face is a discrete random variable.
Probability Distribution of a Discrete Random Variable
The probability distribution of a discrete random variable lists:
- The possible values of the variable
- The probability associated with each value
\( P(X = x_i) \ge 0 \quad \text{for all } x_i \)
\( \sum P(X = x_i) = 1 \)
Each probability must be between 0 and 1, and the sum of all probabilities must equal 1.
Example:
A fair 6-sided die is rolled. Let \( X \) be the random variable representing the number that appears. Give the probability distribution of \( X \).
▶️ Answer/Explanation
Possible values of X: 1, 2, 3, 4, 5, 6
Probability for each value:
Since the die is fair:
\( P(X = x) = \frac{1}{6} \quad \text{for } x = 1, 2, 3, 4, 5, 6 \)
Probability distribution table:
| X | P(X = x) |
|---|---|
| 1 | 1/6 |
| 2 | 1/6 |
| 3 | 1/6 |
| 4 | 1/6 |
| 5 | 1/6 |
| 6 | 1/6 |
\( \sum P(X=x) = 6 \times \frac{1}{6} = 1 \)
Conclusion: The distribution is valid as all probabilities are between 0 and 1, and they sum to 1.
Expected Value (Mean) for Discrete Data
Expected Value (Mean) for Discrete Data
The expected value or mean of a discrete random variable \( X \) gives the long-run average value of \( X \) after many repetitions of the experiment.
\( E(X) = \mu = \sum x_i P(X = x_i) \)
Where:
- \( x_i \) = possible values of \( X \)
- \( P(X = x_i) \) = probability of \( X = x_i \)
The expected value is a weighted average of all possible values of \( X \).
Example:
A spinner has three sectors labeled 1, 2, and 3. The probability distribution is:
| X | P(X = x) |
|---|---|
| 1 | 0.2 |
| 2 | 0.5 |
| 3 | 0.3 |
Find the expected value \( E(X) \).
▶️ Answer/Explanation
Apply the formula
\( E(X) = (1)(0.2) + (2)(0.5) + (3)(0.3) \)
\( = 0.2 + 1.0 + 0.9 = 2.1 \)
Conclusion: The expected value (mean) is 2.1.
Example:
Consider the following probability distribution:
| x | 10 | 20 | 30 |
|---|---|---|---|
| P(X=x) | a | b | 0.5 |
Given that \( E(X) = 23 \), find the values of \( a \) and \( b \).
▶️ Answer/Explanation
Since total probability = 1:
\( a + b + 0.5 = 1 \)
\( a + b = 0.5 \) …(1)
Since \( E(X) = 23 \):
\( 10a + 20b + 30(0.5) = 23 \)
\( 10a + 20b + 15 = 23 \)
\( 10a + 20b = 8 \) …(2)
From (1): \( a = 0.5 – b \)
Substitute into (2):
\( 10(0.5 – b) + 20b = 8 \)
\( 5 – 10b + 20b = 8 \)
\( 5 + 10b = 8 \)
\( 10b = 3 \)
\( b = 0.3 \)
Then:
\( a = 0.5 – 0.3 = 0.2 \)
Applications of Expected Value (Mean) of a Discrete Random Variable
Applications of Expected Value (Mean) of a Discrete Random Variable
The expected value or mean is widely used in real-life and mathematical contexts where we want to predict the average outcome over many trials.
Common applications include:
- Games of chance: To determine whether a game is fair or what a player can expect to win or lose on average (e.g. dice, cards, lotteries).
- Insurance: To calculate premiums based on the expected cost of claims.
- Business decisions: To assess average profit, cost, or demand over time in uncertain conditions.
- Quality control: To predict average defects or failures in manufactured products.
- Risk analysis: To evaluate expected loss or return in investments.
In all these situations, expected value helps make informed decisions by giving a long-run average, even though actual outcomes vary from trial to trial.
\( E(X) = \sum x_i P(X = x_i) \)
Example:
A game costs $5 to play. You win $20 if you spin a 1 on a spinner (probability 0.1). Otherwise, you win nothing. Find the expected gain.
▶️ Answer/Explanation
Let \( X \) = net gain.
\( X = 20 – 5 = 15 \) (if win)
\( X = -5 \) (if no win)
\( E(X) = (15)(0.1) + (-5)(0.9) \)
\( = 1.5 – 4.5 = -3.0 \)
Conclusion: On average, the player loses $3 per game. This is an example of how expected value assesses fairness.