Classical Probability

Question 1

Probability with equally likely outcomes

Answer 1

|S|= number of elements of S
|A|= number of elements of A
A is a subset of S

P(A) = |A|/|S|

Question 2

Permutations

Answer 2

Suppose now that we have n objects.
There are
n(n − 1)(n − 2) · · · 3 · 2 · 1 = n!
different permutations of the n objects.

Question 3

Combinations

Answer 3

Notation and terminology
We define (n r),
for r < n, by

(n r) = n! / (n − r)! r!

and say that
(n r) –> (read as “n choose r”)
represents the number of possible combinations of n objects taken r at a time.