For this, I’m going to look at the eight rules of probability laid out here and the four different counting methods (see more here).
Eight rules of probability
Rule #1: For any event A, 0 = P(A) = 1; in other words, the probability of an event can range from 0 to 1.
Rule #2: The sum of the probabilities of all possible outcomes always equals 1.
Rule #3: P(not A) = 1 — P(A); This rule explains the relationship between the probability of an event and its complement event. A complement event is one that includes all possible outcomes that aren’t in A.
Rule #4: If A and B are disjoint events (mutually exclusive), then P(A or B) = P(A) + P(B); this is called the addition rule for disjoint events
Rule #5: P(A or B) = P(A) + P(B) — P(A and B); this is called the general addition rule.
Rule #6: If A and B are two independent events, then P(A and B) = P(A) * P(B); this is called the multiplication rule for independent events.
Rule #7: The conditional probability of event B given event A is P(B|A) = P(A and B) / P(A)
Rule #8: For any two events A and B, P(A and B) = P(A) * P(B|A); this is called the general multiplication rule
Counting Methods
Factorial Formula: n! = n x (n -1) x (n — 2) x … x 2 x 1
Use when the number of items is equal to the number of places available.
Eg. Find the total number of ways 5 people can sit in 5 empty seats.
= 5 x 4 x 3 x 2 x 1 = 120
Fundamental Counting Principle (multiplication)
This method should be used when repetitions are allowed and the number of ways to fill an open place is not affected by previous fills.
Eg. There are 3 types of breakfasts, 4 types of lunches, and 5 types of desserts. The total number of combinations is = 5 x 4 x 3 = 60
Permutations: P(n,r)= n! / (n-r)!
This method is used when replacements are not allowed and order of item ranking matters.
Eg. A code has 4 digits in a particular order and the digits range from 0 to 9. How many permutations are there if one digit can only be used once?
P(n,r) = 10!/(10–4)! = (10x9x8x7x6x5x4x3x2x1)/(6x5x4x3x2x1) = 5040
Combinations Formula: C(n,r)=(n!)/[(n-r)!r!]
This is used when replacements are not allowed and the order in which items are ranked does not mater.
Eg. To win the lottery, you must select the 5 correct numbers in any order from 1 to 52. What is the number of possible combinations?
C(n,r) = 52! / (52–5)!5! = 2,598,960
