Welcome to the exciting world of Discrete Random Variables! In this chapter, we are zooming in on a specific type of probability model: the Geometric Distribution.
Don't worry if probability models sometimes feel abstract. The Geometric Distribution is highly intuitive because it models something we do every day: waiting for the first success.
You will learn the specific conditions required to use this model, how to calculate the probability of waiting a certain number of trials, and how to find the average waiting time (the Mean).
1. Understanding the Geometric Distribution \(X \sim \text{Geo}(p)\)
The Geometric distribution is used when we perform a sequence of independent trials, and we are interested in the number of trials needed to achieve the very first success.
The random variable \(X\) is defined as:
\(X\) = The number of trials up to and including the first successful outcome.
Since \(X\) counts trials, its possible values are \(x = 1, 2, 3, \dots\) It is a discrete random variable because you can't have 1.5 trials!
Analogy: The Basketball Free Throw
Imagine a basketball player who reliably makes 30% of their free throws (\(p=0.3\)). We keep watching them shoot until they make one successful shot. The Geometric distribution models how many total shots we have to wait for until the first basket goes in.
- If they succeed on the first shot, \(X=1\).
- If they fail, then succeed, \(X=2\).
- If they fail, fail, then succeed, \(X=3\), and so on.
Key Takeaway: The Geometric distribution is a "waiting time" distribution. It counts trials until the first success.
2. Conditions for Application (When to use Geo)
The geometric distribution is a special case of the Bernoulli trial process. It requires four crucial conditions to be met. You can remember these conditions using the adapted mnemonic SIT:
- Success or Failure (Bernoulli Trials): Each trial must have only two possible outcomes, conventionally called "success" and "failure".
- Independence: The outcome of any trial must not affect the outcome of any subsequent trial.
- Trials until First Success: The experiment must continue indefinitely (theoretically) until the first success occurs. (Note: Unlike the Binomial distribution, there is no fixed number of trials, \(n\)).
- Same Probability (\(p\)): The probability of success, \(p\), must be constant for every single trial.
Quick Check: Geometric vs. Binomial
A common mistake is confusing Geometric and Binomial distributions. Remember the stopping rule:
- Binomial (\(X \sim B(n, p)\)): Stops after a fixed number of trials, \(n\).
- Geometric (\(X \sim \text{Geo}(p)\)): Stops after the first success.
3. Calculating Probabilities \(P(X=x)\)
Let \(p\) be the probability of success, and let \(q = 1-p\) be the probability of failure.
If we want the first success to occur on the \(x\)th trial, this means the first \(x-1\) trials must all be failures, and the \(x\)th trial must be a success.
Since the trials are independent, we multiply the probabilities:
Failure on 1st, 2nd, ..., (x-1)th trial AND Success on the xth trial.
$$P(X=x) = \underbrace{q \times q \times \dots \times q}_{x-1 \text{ times}} \times p$$
The Formula for Specific Probability
If \(X \sim \text{Geo}(p)\), the probability function is:
$$P(X=x) = (1-p)^{x-1} p \quad \text{for } x = 1, 2, 3, \dots$$
Or, using \(q\):
$$P(X=x) = q^{x-1} p$$
Example Calculation (Step-by-Step)
A biased coin has \(P(\text{Heads}) = 0.4\). Let \(X\) be the number of tosses until the first Head appears. \(X \sim \text{Geo}(0.4)\). Find the probability that the first Head occurs on the 4th toss.
Step 1: Identify \(p\) and \(q\).
\(p = 0.4\) (Success: Heads).
\(q = 1 - 0.4 = 0.6\) (Failure: Tails).
Step 2: Define the event.
We want \(X=4\). This means the sequence must be (Failure, Failure, Failure, Success).
Step 3: Apply the formula.
$$P(X=4) = q^{4-1} p = q^3 p$$
$$P(X=4) = (0.6)^3 \times (0.4)$$
$$P(X=4) = 0.216 \times 0.4 = 0.0864$$
Cumulative and Tail Probabilities
Sometimes you need to find the probability that the wait is longer than a certain number of trials. This is often the simplest calculation in Geometric distribution!
If we want to find \(P(X > k)\), this means the first success did NOT happen in the first \(k\) trials. In other words, the first \(k\) trials were all failures.
$$P(X > k) = P(\text{1st fail} \cap \text{2nd fail} \cap \dots \cap k\text{th fail})$$
$$P(X > k) = q \times q \times \dots \times q \quad (k \text{ times})$$
The Tail Probability Formula
$$P(X > k) = (1-p)^k = q^k$$
Example: Using the coin from above (\(p=0.4\), \(q=0.6\)), find the probability that it takes more than 5 tosses to get the first Head.
$$P(X > 5) = q^5 = (0.6)^5 = 0.07776$$
If you need \(P(X \le k)\) (the cumulative probability), you can use the complement rule:
$$P(X \le k) = 1 - P(X > k) = 1 - q^k$$
Key Takeaway: The probability structure is simple: failures multiplied by the final success. The formula \(P(X > k) = q^k\) is a powerful shortcut.
4. Mean and Variance of \(X \sim \text{Geo}(p)\)
In any random variable context, the Mean (\(E(X)\) or \(\mu\)) tells us the expected average outcome, and the Variance (\(Var(X)\) or \(\sigma^2\)) tells us how spread out the results are.
The syllabus requires both the knowledge of these results and awareness that their derivations are expected (typically using Probability Generating Functions, which are covered in FS1.4).
4.1 The Mean (Expected Waiting Time)
The expected number of trials until the first success is simply the reciprocal of the probability of success.
$$E(X) = \mu = \frac{1}{p}$$
Conceptual Check: This makes perfect sense! If a bus arrives, on average, 1 out of every 10 minutes (\(p=0.1\)), you would expect to wait 10 minutes on average until the next one arrives (\(1/0.1 = 10\)).
Example: If \(p=0.25\), then the expected number of trials is \(E(X) = 1/0.25 = 4\). You expect to wait 4 trials on average.
4.2 The Variance
The variance measures the spread or variability in the waiting time. If \(p\) is very small (it's hard to succeed), the variance will be high, meaning the waiting time is highly unpredictable.
$$Var(X) = \sigma^2 = \frac{1-p}{p^2} = \frac{q}{p^2}$$
Did you know? As \(p\) increases (the success becomes easier), the mean (\(1/p\)) decreases, and the variance (\(q/p^2\)) also decreases, meaning the outcome becomes more consistent and happens sooner.
Summary of Key Formulas for Geo(\(p\))
Probability: \(P(X=x) = q^{x-1} p\)
Tail Probability: \(P(X > k) = q^k\)
Mean: \(E(X) = \frac{1}{p}\)
Variance: \(Var(X) = \frac{q}{p^2}\)
5. Common Pitfalls and Memory Aids
Mistake 1: Miscounting \(x\)
Students sometimes confuse whether \(X\) is the number of failures *before* the first success, or the number of trials *up to and including* the first success.
Remember: In OxfordAQA Further Maths (9665), \(X\) is defined as the total number of trials, so \(x \ge 1\). If the question defines the variable differently (e.g., number of failures before success), you must adjust the formula accordingly, but generally stick to \(X \sim \text{Geo}(p)\) where \(x\) starts at 1.
Mistake 2: Forgetting Independence
If the probability of success changes based on previous outcomes (e.g., drawing cards without replacement), the Geometric model cannot be used because the trials are not independent.
Memory Aid: Check the SIT conditions thoroughly before applying the formula.
Example: Finding $p$ from the Mean
If you are told that the expected number of attempts needed to pass an exam is 5, you can immediately find \(p\):
$$E(X) = 5$$ $$\frac{1}{p} = 5$$ $$p = \frac{1}{5} = 0.2$$
This is a fast way to solve certain exam questions!
Congratulations, you now have a solid understanding of the Geometric Distribution. Keep practicing those calculations, and you'll master this topic!