Mathematics · 21 / 23
Binomial model: count successes
Final-year concepts · Worked example and 4 exercises with solutions, at your pace.
The principle, simply
A binomial model requires a fixed number of independent trials, two outcomes and constant success probability p. X counts successes, not their positions. For exactly k successes in n trials, count placements and multiply by pᵏ(1−p)ⁿ⁻ᵏ. Its mean np is not a guaranteed outcome.
Multiply along a path; add disjoint paths for one event. This partial tree omits branches after not A.
- 01Check trials
- 02Define X
- 03Count placements
- 04Use complements
WORKED EXAMPLE
From principle to reasoning
With three independent trials and p = 0.2, one success has three possible positions: 3 × 0.2 × 0.8² = 0.384.
Solve a problem in this subject
- Define the objects
State unknowns, domain and data. Real square roots need nonnegative arguments; denominators must be nonzero.
- Choose a transformation
Name the rule: expansion, equivalent equations, factoring, differentiation. Check its conditions. Squaring an equation requires testing candidates in the original.
- Show the reasoning
Each line should follow from the previous one. Keep exact values where possible and distinguish equality, approximation and implication.
- Test and conclude
Substitute into the original expression. Check graph points and domain boundaries. State a solution set or the required unit.
YOUR TURN
01 · Try before checking
Two independent trials with p = 0.3: find the probability of at least one success.
One hint
First find the probability of no successes.
Reveal and understand the solution
P(X ≥ 1) = 1 − 0.7² = 0.51. No successes and at least one are complementary events.
Compare method, conditions and answer; identify the exact part to revisit.
IMPROVE THROUGH PRACTICE
Three additional challenges with solutions
Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.
When can five coin tosses be modelled as binomial?
A hint to get started
Check fixed count, two outcomes, independence and constant probability.
Step-by-step solution
- Fix five tosses and define heads as success.
- Assume independence and constant P(heads).
- The head count is B(5,p), with p = 0.5 for a fair coin.
Which step did you miss? Retry without help, then explain why the method works.
X follows B(4,0.5). Find P(X = 2).
A hint to get started
There are six ways to place two successes in four trials.
Step-by-step solution
- P(X = 2) = 6 × 0.5² × 0.5².
- Each sequence has probability 1/16.
- The total is 6/16 = 0.375.
Which step did you miss? Retry without help, then explain why the method works.
For X distributed B(10,0.2), find P(X ≥ 1) and its expectation.
A hint to get started
At least one is the complement of zero.
Step-by-step solution
- P(X = 0) = 0.8¹⁰.
- P(X ≥ 1) = 1 − 0.8¹⁰ ≈ 0.8926.
- E(X) = np = 2; it is not a certain next outcome.
Which step did you miss? Retry without help, then explain why the method works.
The trap to avoid
Sampling without replacement usually changes probabilities and breaks independence.
Without notes, can you…
- Explain the principle in your own words?
- Rebuild the diagram and its connections?
- Solve the exercise and justify the method?
Choosing a state saves it in this browser. This is your assessment, not a grade.