CIE AS/A Level Maths-5.4 Discrete random variables- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-5.4 Discrete random variables- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-5.4 Discrete random variables- Study Notes
Key Concepts:
- Probability Distribution of a Discrete Random Variable
- Binomial and Geometric Distributions
- Expectation and Variance for Binomial and Geometric Distributions
Probability Distribution of a Discrete Random Variable
Probability Distribution of a Discrete Random Variable
A discrete random variable \(X\) is one which takes only certain distinct values. The probability distribution of \(X\) shows all possible values of \(X\) and their corresponding probabilities.
Properties:
- \(0 \leq P(X = x_i) \leq 1\) for all \(i\)
- \(\sum P(X = x_i) = 1\)
Expectation (Mean):
\(E(X) = \sum \big( x_i \cdot P(X = x_i) \big)\)
Variance:
\(\text{Var}(X) = E(X^2) – \big(E(X)\big)^2\)
Example :
A fair coin is tossed twice. Let \(X\) = number of heads obtained.
Construct the probability distribution of \(X\), and find \(E(X)\) and \(\text{Var}(X)\).
▶️ Answer/Explanation
Possible outcomes: {HH, HT, TH, TT}
Values of \(X\):
- \(X = 0\) (TT) → Probability = 1/4
- \(X = 1\) (HT, TH) → Probability = 2/4 = 1/2
- \(X = 2\) (HH) → Probability = 1/4
Probability Distribution Table:
| \(X\) | 0 | 1 | 2 |
| \(P(X)\) | \(\dfrac{1}{4}\) | \(\dfrac{1}{2}\) | \(\dfrac{1}{4}\) |
Step 1: Find \(E(X)\)
\(E(X) = 0 \cdot \dfrac{1}{4} + 1 \cdot \dfrac{1}{2} + 2 \cdot \dfrac{1}{4}\)
\(E(X) = 0 + \dfrac{1}{2} + \dfrac{1}{2} = 1\)
Step 2: Find \(E(X^2)\)
\(E(X^2) = 0^2 \cdot \dfrac{1}{4} + 1^2 \cdot \dfrac{1}{2} + 2^2 \cdot \dfrac{1}{4}\)
\(E(X^2) = 0 + \dfrac{1}{2} + 1 = \dfrac{3}{2}\)
Step 3: Find Variance
\(\text{Var}(X) = E(X^2) – \big(E(X)\big)^2\)
\(\text{Var}(X) = \dfrac{3}{2} – (1)^2 = \dfrac{1}{2}\)
Final Answer: \(E(X) = \boxed{1}, \; \text{Var}(X) = \boxed{\dfrac{1}{2}}\)
Binomial and Geometric Distributions
Binomial Distribution
A random variable \(X\) follows a binomial distribution if it counts the number of successes in \(n\) independent trials, where each trial has only two outcomes: success or failure.
Notation:
\(X \sim B(n, p)\), where:
- \(n\) = number of trials
- \(p\) = probability of success on a single trial
Probability mass function (pmf):
\( P(X = k) = \binom{n}{k} p^k (1-p)^{\,n-k}, \quad k = 0, 1, 2, \dots, n \)
Mean and variance:
\( E(X) = np, \quad \text{Var}(X) = np(1-p) \)
- Practical situations: number of heads in \(n\) coin tosses, number of defective items in a batch, number of successes in repeated experiments with fixed probability.
Geometric Distribution
A random variable \(Y\) follows a geometric distribution if it counts the number of trials until the first success occurs.
Notation:
\(Y \sim \text{Geo}(p)\), where \(p\) = probability of success on each trial.
Probability mass function (pmf):
\( P(Y = r) = p(1-p)^{\,r-1}, \quad r = 1, 2, 3, \dots \)
Mean and variance:
\( E(Y) = \dfrac{1}{p}, \quad \text{Var}(Y) = \dfrac{1-p}{p^2} \)
- Practical situations: number of tosses until the first head, number of attempts to achieve first success, waiting time for an event in repeated trials.
Example:
A coin is tossed 5 times. Find the probability of getting exactly 3 heads. Also, find the expected number of heads.
▶️ Answer/Explanation
Here \(X \sim B(5, 0.5)\), since a fair coin is used.
Probability of exactly 3 heads:
\( P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^2 = 10 \cdot 0.125 \cdot 0.25 = 0.3125 \)
Expected number of heads: \( E(X) = np = 5 \cdot 0.5 = 2.5 \)
Final Answer: \( P(X=3) = \boxed{0.3125}, \quad E(X) = \boxed{2.5} \)
Example:
A die is rolled repeatedly until a 6 appears. Find the probability that the first 6 appears on the 4th roll, and the expected number of rolls until the first 6.
▶️ Answer/Explanation
Let \(Y \sim \text{Geo}(p)\), with \(p = 1/6\).
Probability that first 6 appears on 4th roll:
\( P(Y = 4) = p(1-p)^{4-1} = \dfrac{1}{6} \left(\dfrac{5}{6}\right)^3 = \dfrac{125}{1296} \approx 0.09645 \)
Expected number of rolls: \( E(Y) = \dfrac{1}{p} = 6 \)
Final Answer: \( P(Y=4) = \boxed{\dfrac{125}{1296}}, \quad E(Y) = \boxed{6} \)
Expectation and Variance for Binomial and Geometric Distributions
Binomial Distribution \(X \sim B(n, p)\)
Expectation (mean):
\( E(X) = np \)
Variance:
\( \text{Var}(X) = np(1-p) \)
Standard deviation:
\( \sigma_X = \sqrt{\text{Var}(X)} = \sqrt{np(1-p)} \)
Use: Calculate average number of successes and spread around the mean in repeated independent trials.
Geometric Distribution \(Y \sim \text{Geo}(p)\)
Expectation (mean):
\( E(Y) = \dfrac{1}{p} \)
Variance:
\( \text{Var}(Y) = \dfrac{1-p}{p^2} \)
Use: Calculate the average number of trials until the first success and spread.
Example:
A fair coin is tossed 10 times. Find the expected number of heads and the variance.
▶️ Answer/Explanation
Here \( X \sim B(10, 0.5) \)
Expectation: \( E(X) = np = 10 \cdot 0.5 = 5 \)
Variance: \( \text{Var}(X) = np(1-p) = 10 \cdot 0.5 \cdot 0.5 = 2.5 \)
Final Answer: \( E(X) = \boxed{5}, \; \text{Var}(X) = \boxed{2.5} \)
Example:
A die is rolled repeatedly until a 6 appears. Find the expected number of rolls and the variance.
▶️ Answer/Explanation
Here \( Y \sim \text{Geo}(p) \), with \( p = 1/6 \)
Expectation: \( E(Y) = \dfrac{1}{p} = \dfrac{1}{1/6} = 6 \)
Variance: \( \text{Var}(Y) = \dfrac{1-p}{p^2} = \dfrac{5/6}{(1/6)^2} = 30 \)
Final Answer: \( E(Y) = \boxed{6}, \; \text{Var}(Y) = \boxed{30} \)