germany-abitur

Papers (25)
2025
abitur__nrw_analysis-gk 3 abitur_nrw__typ_analysis_lk 4 abitur_nrw__typ_stochastik_gk 3 abitur_nrw__typ_stochastik_lk 3
2024
abitur__bayern_geometrie 9 abitur__bayern_infinitesimalrechnung 2 abitur__bayern_stochastik 9
2023
abitur__bayern_geometrie 9 abitur__bayern_infinitesimalrechnung 2 abitur__bayern_stochastik 10
2022
abitur__bayern_geometrie 9 abitur__bayern_infinitesimalrechnung 1 abitur__bayern_stochastik 8
2021
abitur__bayern_geometrie 10 abitur__bayern_infinitesimalrechnung 2 abitur__bayern_stochastik 9
2020
abitur__bayern_geometrie 7 abitur__bayern_infinitesimalrechnung 2 abitur__bayern_stochastik 9
2019
abitur__bayern_geometrie 9 abitur__bayern_infinitesimalrechnung 2 abitur__bayern_stochastik 2
2018
abitur__bayern_geometrie 1 abitur__bayern_infinitesimalrechnung 2 abitur__bayern_stochastik 2
2024 abitur__bayern_stochastik

9 maths questions

QA a 3 marks Probability Definitions Finite Equally-Likely Probability Computation View
Show by calculation that the probability of winning the game is $\frac { 1 } { 4 }$.
QA b 2 marks Binomial Distribution MCQ Selecting a Binomial Probability Expression or Value View
The game is played five times. Give an event in the context of the problem whose probability can be calculated using the term $\left( \frac { 1 } { 4 } \right) ^ { 2 } \cdot \left( \frac { 3 } { 4 } \right) ^ { 3 }$.
QB 1a 3 marks Approximating Binomial to Normal Distribution View
Determine the probability that among 200 randomly selected packages, more than one quarter are returns.
QB 1b 3 marks Binomial Distribution MCQ Selecting a Binomial Probability Expression or Value View
Describe in the context of the problem a random experiment in which the probability of an event can be calculated using the term $1 - \sum _ { i = 0 } ^ { 8 } \binom { 30 } { i } \cdot 0,2 ^ { i } \cdot 0,8 ^ { 30 - i }$, and give this event.
QB 1c 4 marks Binomial Distribution Find Minimum n for a Probability Threshold View
Determine how many packages must be randomly selected at minimum so that the probability that at least one return is among them is greater than $90 \%$.
QB 1d 5 marks Conditional Probability Total Probability via Tree Diagram (Two-Stage Partition) View
$49 \%$ of the packages contain clothing. Of the packages that are returns, $91 \%$ contain clothing.
A package is randomly selected. The following events are considered: $R$ : ``The package is a return.'' $K$ : ``The package contains clothing.'' Represent the described situation in a completely filled four-field table. For the case that the selected package is not a return, determine the probability that the package contains clothing.
QB 1e 2 marks Permutations & Arrangements Probability via Permutation Counting View
Determine the probability that the first two packages are returns.
QB 1f 4 marks Binomial Distribution Compute Cumulative or Complement Binomial Probability View
Determine the probability that two or three returns are removed.
QB 2 4 marks Discrete Probability Distributions Probability Distribution Table Completion and Expectation Calculation View
The table shows the probability distribution of a random variable X, which can only take the values $1,2,3,4$ and 5.
k12345
$\mathrm { P } ( \mathrm { X } = \mathrm { k } )$$\mathrm { p } _ { 1 }$$\mathrm { p } _ { 2 }$$\mathrm { p } _ { 3 }$0,20,15

The probabilities $P ( X = 4 )$ and $P ( X = 5 )$ as well as the expected value and the variance of X are known. From this information, the following system of equations results, with which the missing probabilities $p _ { 1 } , p _ { 2 }$ and $p _ { 3 }$ can be calculated.
I $p _ { 1 } + p _ { 2 } + p _ { 3 } = 0,65$ II $p _ { 1 } + 2 p _ { 2 } + 3 p _ { 3 } = 1,45$ III $\quad 4 p _ { 1 } + p _ { 2 } = 0,6$ Determine, without solving the system of equations, which values for the expected value and the variance of X were used when setting up the system of equations.