Combinatorics and classical probability
With equally likely outcomes, P = favourable ÷ possible outcomes. Combinatorics helps you count them.
Formulas
- Classical: P(A) = |A| / |Ω|
- No order, no repetition: C(n, k) = n! / (k! · (n − k)!)
- With order, no repetition: n! / (n − k)!
Your turn
Three problems on this topic. Comma or point, fractions like 1/6 work too.
Task
Two fair dice are rolled. What is the probability that the total is exactly 8?
P (4 decimal places)
Comma or point; fractions like 1/6 work too.
Worked example
Two fair dice are rolled. What is the probability that the total is exactly 5?
Worked solution
- 1.Laplace: P = favourable / possible outcomes; possible outcomes: 6 · 6 = 36
- 2.Favourable (total 5): 4 pairs
- 3.P = 4 / 36 = 0.1111
Typical mistake: Counting (2,5) and (5,2) only once; they are two different outcomes.
A working group of 2 is chosen from 11 students. How many different groups are possible?
Worked solution
- 1.No order, no repetition: binomial coefficient
- 2.(11 über 2) = 11! / (2! · 9!) = 55
Typical mistake: Counting with order (n!/(n−k)!); order does not matter in a group.
Frequently asked questions
How do I know if order matters?
If a different order is a different result (places, PINs) it matters; for groups or lottery numbers it does not.