Introduction
Once you have identified that a random variable follows a binomial or geometric distribution, the next natural question is: on average, what value do we expect? And for the binomial: how spread out are the outcomes? These questions are answered by the expectation (mean) and variance formulae — compact, elegant results that allow you to bypass lengthy probability calculations.
In 9709 examinations, these formulae appear regularly in multi-part questions. You will typically be asked to state or use and directly, to solve for an unknown parameter or , or to combine these results with the general properties of expectation and variance. Proofs of the formulae are not required by the syllabus, but you must be able to apply them fluently.
Core Concept
The Binomial Distribution
If , then counts the number of successes in independent trials, each with probability of success (and failure ).
The expectation reflects the intuitive idea that, across trials, you expect a fraction to succeed. The variance captures how much the count fluctuates — it is largest when (maximum uncertainty) and shrinks as approaches 0 or 1.
The Geometric Distribution
If , then counts the number of trials needed to obtain the first success, so .
The expectation is deeply intuitive: if each trial has a chance of success, you expect to wait about 5 trials for the first success.
Syllabus note: The 9709 syllabus requires the expectation formula for the geometric distribution only. The variance formula exists but is outside the scope of this objective — do not spend time on it unless a question explicitly provides it.
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