Repeat. Count.: Toss a coin exactly 10 times. Count the heads. The binomial distribution tells us the chance of each result: 0 heads, 1 head, …, or 10 heads.
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What does “success” mean?
The event you choose to count.
Heads example: heads = success; tails = failure.
Faulty items: if X counts faulty items, a fault is a “success”.
What does “independent” mean?
One result does not affect the next.
Coin example: after a head, the chance of another head is still 0.5.
What does the distribution tell us?
The probability of each possible count of successes.
Example: P(X = 3) means the chance of exactly 3 heads in all 10 tosses.
Try another example
A sensor is tested 12 times independently. It detects movement with probability 0.8 each time. X counts detections. State the model.
Step by step
- What are we counting? Detections.
- How many tests? n = 12.
- Chance of detection on one test? p = 0.8.
Final answer
X ~ B(12, 0.8).
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Same experiment, repeated a fixed number of times: Count how often your chosen event happens. Check all four conditions below.
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Exam question: what are we assuming?
A workshop models successful repairs in 25 jobs as binomial. State one assumption.
An assumption is something we treat as true so the model can work.
Answer 1: each job has the same chance of a successful repair.
Or answer 2: success on one job does not change the chance of success on another.
Only one is needed. “25 jobs” is already given; it is not an assumption.
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Expected = average; spread = variation: For X ~ B(20, 0.3), there are 20 attempts and a 0.3 chance of success each time.
| Find | Rule | Substitute |
|---|---|---|
| Mean / expected number | np | 20 × 0.3 = 6 |
| Variance | np(1 − p) | 20 × 0.3 × 0.7 = 4.2 |
| Standard deviation | √variance | √4.2 ≈ 2.05 |
Does “expected 6” guarantee 6 successes?
No. It means 6 on average over many groups of 20 attempts. One group can have a different count.
Why use 0.7 for the variance?
p = 0.3, so 1 − p = 0.7 is the chance of failure.
Expected profit example
80 calls, chance of a sale = 0.2.
Expected sales: 80 × 0.2 = 16.
Each sale earns $15; daily cost is $200.
Expected profit: 16 × 15 − 200 = $40.
Working backwards: start with np: Expected number = attempts × chance of success. Put the limit on this average.
Maximum bookings
54 seats. Each booked passenger arrives with probability 0.9. Expected arrivals must be at most 54. Find the maximum bookings.
Step by step
- Write the limit: 0.9n ≤ 54.
- Solve: n ≤ 54 ÷ 0.9 = 60.
- Check the boundary: 60 bookings → 54 expected arrivals; 61 → 54.9, too many.
Final answer
Maximum: 60 bookings.
What if the answer is a decimal?
0.85n ≤ 50 → n ≤ 58.8235…
Choose 58, since 59 is above the limit. Check it in the original condition.
What if p is missing?
40 attempts; expected successes = 12.
40p = 12 → p = 12 ÷ 40 = 0.3.
Does this guarantee enough seats?
No. This question limits the average arrivals. A probability of overcrowding is a different question.
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Calculating P(X = r)
The meaning: C(n,r) = ways to arrange r successes in n trials. pr = prob of r successes.
(1-p)n-r = prob of n-r failures.
Calculate a probability
B(10, 0.5): Find P(X = 3).
Step by step
- Use formula: P(X=3) = C(10,3) × 0.53 × 0.57
- C(10,3) = 10!/(3! × 7!) = 120
- P(X=3) = 120 × (0.5)10 = 120/1024 ≈ 0.117
Final answer
P(X = 3) ≈ 0.117 or 11.7%.