Abitur · 10% of test plan
Mathematik — Stochastik (Statistics & Probability) for the Abitur Exam
Stochastik is the second major block in Abitur Mathematik, accounting for approximately 25–30% of the exam. It covers: classical probability (Laplace), combinatorics (Binomialkoeffizient), Binomialverteilung (binomial distribution), and hypothesis testing (Signifikanztests). Applied probability problems from real-world contexts are standard.
Locale-specific study guides
Pass-rate data, regulatory context, and study tips for Mathematik — Stochastik (Statistics & Probability) all change by candidate locale. Pick your context:
- Mathematik — Stochastik (Statistics & Probability) · United StatesCalibrated for American candidates
- Mathematik — Stochastik (Statistics & Probability) · United KingdomCalibrated for British candidates
- Mathematik — Stochastik (Statistics & Probability) · IndiaCalibrated for Indian candidates
- Mathematik — Stochastik (Statistics & Probability) · PhilippinesCalibrated for Filipino candidates
- Mathematik — Stochastik (Statistics & Probability) · NigeriaCalibrated for Nigerian candidates
Common failure modes
These are the patterns that cause most candidates to lose marks on this topic. Recognising them in advance is half the work.
- !Confusing "with replacement" and "without replacement" in probability calculations
- !Incorrect binomial coefficient calculation: n over k (n!/(k!(n-k)!))
- !Not correctly formulating the null hypothesis for significance tests
Study tips
- 1Memorize the binomial probability formula: P(X = k) = C(n,k) × p^k × (1-p)^(n-k).
- 2Practice cumulative binomial probability tables — many Abitur questions ask for P(X ≤ k) or P(X ≥ k).
- 3For hypothesis tests, follow the standard steps: H₀ and H₁ formulation → significance level → critical region → decision.
Sample Abitur Mathematik — Stochastik (Statistics & Probability) questions
These sample items mirror the format and difficulty of real Abitur questions. Practice with thousands more on the free Koydo question bank.
- 1
Ein fairer Würfel wird 6-mal geworfen. Mit welcher Formel berechnet man P(genau 2-mal eine 6)? (A fair die is rolled 6 times. Which formula gives P(exactly 2 sixes)?)
- AP = C(6,2) × (1/6)² × (5/6)⁴Correct
- BP = 2 × (1/6) × (5/6)
- CP = (1/6)² × (5/6)⁴
- DP = C(6,2) × (1/6)⁶
Why this answer?
The binomial formula is B(n,k,p): P(X=k) = C(n,k) × p^k × (1-p)^(n-k). Here n=6, k=2, p=1/6: P = C(6,2) × (1/6)² × (5/6)⁴. C(6,2) = 15; (1/6)² = 1/36; (5/6)⁴ = 625/1296.
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