Fixed trials, two outcomes, constant p: Use the binomial model X ~ B(n, p) when there is a fixed number n of independent trials, each with just two outcomes (success/failure) and the same success probability p every time.
X = the number of successes.
The binomial distribution: bars for X = 0, 1, …, n; change n and p to watch the shape shift.
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IB-style question — set it up
A biased coin lands heads with probability 0.4.
It is tossed 12 times.
State the distribution of the number of heads X.
Step by step
- Fixed trials n, constant p, two outcomes.
- Write the model.
Final answer
X ~ B(12, 0.4).
p must stay constant: If the probability changes between trials (e.g. drawing without replacement), it is not binomial.
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Exactly k successes: For exactly k successes use binompdf(n, p, k).
By hand it is ⁿCₖ pᵏ (1−p)ⁿ⁻ᵏ, but on Paper 2 the calculator does it directly.
IB-style question — exactly k
X ~ B(10, 0.3).
Find P(X = 4).
Step by step
- Exactly 4 successes → binompdf.
- Evaluate on the GDC.
Final answer
P(X = 4) ≈ 0.200.
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binomcdf adds up from 0: binomcdf(n, p, k) gives P(X ≤ k) (cumulative).
Build the others from it: P(X ≥ k) = 1 − P(X ≤ k−1), and P(a ≤ X ≤ b) = P(X ≤ b) − P(X ≤ a−1).
IB-style question — at most / at least
X ~ B(10, 0.3).
Find P(X ≤ 2) and P(X ≥ 3).
Step by step
- At most 2 → binomcdf.
- At least 3 is the complement.
Final answer
P(X ≤ 2) ≈ 0.383; P(X ≥ 3) ≈ 0.617.
Mind the −1: For P(X ≥ k) subtract P(X ≤ k − 1) — using P(X ≤ k) drops the value k itself.
IB-style question — conditional in a binomial
X ~ B(10, 0.4). Find P(X ≥ 5 | X ≥ 3).
Step by step
- Conditional probability: the X ≥ 5 case sits inside the X ≥ 3 case, so the numerator is just P(X ≥ 5).
- Each from 1 − binomialcdf on the GDC.
Final answer
≈ 0.441.
Complement, and solving for n: P(at least one) = 1 − P(X = 0).
To find n so that P(at least one) reaches a target, solve 1 − (1−p)ⁿ ≥ target (take logs, or use a table on the GDC).
IB-style question — find the number of trials
A machine produces an item that is faulty with probability 0.1.
How many items must be inspected so that the probability of finding at least one faulty item is at least 0.9?
Step by step
- P(at least one) = 1 − P(none) ≥ 0.9.
- Take logs and solve.
Final answer
At least 22 items must be inspected.
Round UP for 'at least': Solving gives a decimal; since n is a whole number of trials and you need to reach the target, round up.