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9709 ~ Geometric Distribution Simplified

Master 9709 S1 Geometric Distribution: key formulas, examples & free download worksheet for Discrete Random Variables. Scholarz Gate makes it simple.
Geometric Distribution Friendly Explainer
Syllabus Extract (2023-2025)

Cambridge AS & A Level Mathematics 9709 (S1)
5.4 Discrete random variables: Geometric distribution
• Use notation Geo(p).
• Apply formula \(P(X=r) = p(1-p)^{r-1}\), \(r = 1, 2, 3, \dots\).
• Recognise suitable contexts: independent trials, constant p, first success.
• Calculate probabilities for exact and cumulative cases.
• Use expectation formula \(E(X) = 1/p\) (proof not required).

Intro: The “First Success” Formula You’ll Actually Remember

Let’s cut through the fog. Geometric distribution isn’t all about memorizing formulas ~ it’s about understanding the rhythm of randomness. You’re running independent trials, each with the same chance of success. The question is: how long till you finally win?

That’s where the magic of \(P(X=r) = p(1-p)^{r-1}\) kicks in. It’s not just a rule ~ it’s a replay of every try.

In this explainer, we break down:
• Why the domain starts at 1 (not 0 ~ don’t fall for that trap)
• How to flip “at least” into complement form like a pro
• What makes geometric “memoryless” ~ and why that’s a superpower
• Lots of practice questions, from CIE past papers and specimens, from the latest in 2025 going back.

Trust me, when you're done with the worksheet below, you won’t just solve geometric problems ~ you’ll see them coming

Crucial Revision Notes
• Always use \(P(X=r) = p(1-p)^{r-1}\) when "first success on the rth trial ".
• Domain always starts at trial 1, never 0.
• \(E(X) = 1/p\). If \(E(X)\) given, solve for \(p\).
• “At least / greater than” → use complement rules:
\(P(X>r) = q^{\,r}\) and \(P(X \le r) = 1-q^{r}\) where \(q=1-p\)
• Geometric is memoryless: previous failures don’t change future.
Question Types Examined
1. Exact trial (1st success on k)

Example: A coin with p=0.3. Find probability first head on trial 4.

Explain This

1. Identify: Want success on trial 4.

2. Apply: \(P(X=4) = (1-0.3)^3 \cdot 0.3 = 0.103\).

3. Final Answer: 0.103 (3 s.f.).

2. At most / At least / Inequalities

Example: p=0.25. Find probability first success within 3 trials.

Explain This

1. “At most 3” = \(X \le 3\).

2. Compute: \(P(X=1)+P(X=2)+P(X=3) = 0.25+0.1875+0.1406=0.578\).

3. Expectation / Finding p

Example: If expected number of trials=5, find p.

Explain This

1. Recall \(E(X)=1/p\).

2. So p=1/5=0.2.

4. Conditional / Memoryless property

Example: Failures in first 2 trials, find probability first success on 5th trial (p=0.3).

Explain This

1. After 2 failures, distribution restarts.

2. Equivalent to “success on 3rd trial of fresh experiment.”

3. \(P(X=3)=(0.7)^2 \cdot 0.3 = 0.147\).

5. Word Problem / Contextual

Example: Machine has p=0.8 success. Probability it fails 3 times before working?

Explain This

1. This is “first success on trial 4.”

2. \(P(X=4)=(0.2)^3 \cdot 0.8=0.0064\).

Practice Worksheet
Download Worksheet Worked Solutions