Sum of two dice (classical probability)
Two dice have 36 equally likely outcomes; (2,5) and (5,2) count as two different outcomes.
Worked example
Two fair dice are rolled. What is the probability that the total is exactly 9?
Worked solution
- 1.Laplace: P = favourable / possible outcomes; possible outcomes: 6 · 6 = 36
- 2.Favourable (total 9): 4 pairs
- 3.P = 4 / 36 = 0.1111
Typical mistake: Counting (2,5) and (5,2) only once; they are two different outcomes.
Your turn
Three problems of exactly this type, each time with new numbers. Comma or point; fractions like 1/6 work too.
Task
Two fair dice are rolled. What is the probability that the total is exactly 11?
P (4 decimal places)
Comma or point; fractions like 1/6 work too.
Formulas for this topic
- Classical: P(A) = |A| / |Ω|
- No order, no repetition: C(n, k) = n! / (k! · (n − k)!)
- With order, no repetition: n! / (n − k)!
Typical mistake: Counting (2,5) and (5,2) only once; they are two different outcomes.
Frequently asked questions
What is the typical mistake in "Sum of two dice (classical probability)"?
Counting (2,5) and (5,2) only once; they are two different outcomes.
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.