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Decide when a count can be modelled by a binomial distribution, define success and identify the fixed-trial, common-probability and independence assumptions.
Before you startDiscrete random variables, independence and probability trees.
01 / Define a success count
Write X~B(n,p).
n is the number of trials; p is the probability of the chosen success on each trial.
Here “success” is just the event being counted. A faulty item, a late parcel or a missed shot can be called a success mathematically. The name does not mean a desirable outcome.
Twelve independent checks each succeed with probability 0.2. X counts successes.
A binomial model is justified: X~B(12,0.2).
02 / Check four features
The count has values 0 through n. The parameter n is a nonnegative integer and 0≤p≤1. Cases p=0 or p=1 are valid degenerate distributions.
Eighteen independent components each have probability 0.04 of failing inspection. X counts failures. State a model.
Failure is the event counted as success.
X~B(18,0.04), under the stated independence and equal-probability assumptions.
For that experiment, Y counts components passing inspection. State a model and its relation to X.
Every component either passes or fails.
Y~B(18,0.96), and Y=18−X. X and Y are not independent of each other.
03 / Two outcomes means two classes for your question
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Success=blue, p=0.5
Failure=red or green, probability 0.5.
There are two classes for this variable
The three raw colours do not prevent a binomial model.
X~B(16,0.5)
Assume spins are independent and the probabilities stay fixed.
For the same spinner, Z counts red or green in 16 spins. State a model.
Add the two mutually exclusive colour probabilities.
Z~B(16,0.5), since 0.2+0.3=0.5.
Could the total of 16 fair die scores be modelled as B(16,1/6)?
A score total is not a success count.
No. The number of sixes could be B(16,1/6) for independent rolls, but the sum of scores has different possible values and probabilities.
04 / Check whether the number of trials is fixed
Trials succeed independently with probability 0.3. X is the number of attempts until the first success. Is X binomial?
What does X count, and is there a fixed trial total?
No. X counts attempts used under a stopping rule, not successes in a fixed number of trials.
At most 6 attempts are allowed, stopping at the first success. Is the attempts-used variable B(6,0.3)?
Six is a cap, not necessarily the actual number run.
No. The number made can vary. Its endpoint mass must account for reaching the cap.
Thirty attempts are always made. X counts successes, each independent with chance 0.3. Is a binomial model appropriate?
Now all four conditions are given.
Yes. X~B(30,0.3).
05 / A common p needs justification
Ten independent checks have success chances 0.1 for the first five and0.4 for the remaining five. Is the total B(10,0.25)?
An average success chance does not preserve the full distribution.
No. The trial probabilities differ. The mean chance 0.25 is not enough to justify a binomial model for the total.
A learner’s chance of solving each successive question increases with practice. What binomial condition is doubtful?
Look at the common-probability assumption.
The success probability may change. Outcomes may also be linked through shared knowledge, so independence needs separate consideration.
06 / Check links between trials
A bag has 4 marked and 6 unmarked tokens. Select 5 without replacement. Is the marked count exactly B(5,0.4)?
The composition changes after each selection.
No. The conditional chance changes, so the selections are dependent. Use a model that accounts for sampling without replacement.
Why might a binomial approximation be useful for a small random sample from a very large population?
Removing a few items may barely change the proportions.
If the sample is small relative to the population, the changing conditional probabilities may be negligible. State this approximation and also check the sampling process; “large” alone does not guarantee independence.
Items have the same individual defect probability, but a shared batch fault can affect many together. Is common p enough?
A shared influence can correlate outcomes.
No. The independence assumption may fail even when all marginal defect probabilities are equal.
07 / State the model in context
Let X be the number of late parcels among the 25 observed
This identifies the count and the observation unit.
Assume a constant late probability and independent parcel outcomes
In real data, common transport disruptions might challenge these assumptions.
Then X~B(25,0.08)
The model is conditional on those assumptions, not a verified claim about a real courier.
If 12 trials all have success probability 0, what is the success count distribution?
No success can occur.
B(12,0) has P(X=0)=1.
If 12 trials all have success probability 1, what is the distribution?
Every trial succeeds.
B(12,1) has P(X=12)=1.
08 / Define, justify and then calculate
Say what X counts, identify a fixed n, define success and its common probability p, and explain independence. Distinguish exact assumptions from approximations. Several raw outcomes are fine if each trial can be classified into the chosen success and its complement.
Section 1 of 8 · Define a success count