Probability: Rules & Terms


Terms, Definitions, and Notation

 

Probability Notation

P(A) is the probability of the event A occurring

P(B) is the probability of the event B occurring

 

Complement

The complement of an event is all outcomes that are not the event. For example, if the given event is being on time, then its complement is being late.

 

Notation:

Ac is the complement of A

Bc is the complement of B

 

Recall that P(S), where S represents the sample space, equals 1. Hence, the probability of an event and its complement, which together create an entire sample space, must also equal 1. 

P(A)+P(Ac)=1

P(B)+P(Bc)=1

 

Mutually Exclusive Events

Events that are mutually exclusive will never occur at the same time. This is synonymous with saying that two events are "disjoint."

 

Independent Events

P(A) and P(B) have no effect on each other, meaning that the probability of event A happening does not affect the probability of event B happening, or vice versa.

 

Conditional Probability

This means that the probability of event A happening affects the probability of a second event B happening.

 

P(A | B) = Probability of A given B

P(B | A) = Probability of B given A

 

The Probability of A "or" B

 P(A begin mathsize 10px style large union end style B) = P(A or B)

 

The Probability of A "and" B

 P(A begin mathsize 10px style large intersection end style B) = P(A and B)

 

Rules

 

The Multiplication Rule

P(A and B) = P(A) \(\times\) P(B) 

  • This is ONLY true for events that are independent.
  • If two events are mutually exclusive/ disjoint, then they could not occur at the same time and the P(A and B) woud equal 0.

The Addition Rule

P(A or B) = P(A) + P(B) - P(A and B)

  • If mutually exclusive/ disjoint, then P(A and B) = 0 and thus P(A or B) = P(A) + P(B)

 

Conditional

\(P(A|B) = {P(A∩B) \over P(B)}\) , provided P(B) > 0 (otherwise, it is not defined)

If the events are independent, then \(P(A|B) = {{P(A)\cdot P(B)}\over P(B)}\)  → the P(B)'s cancel out, leaving you with \(P(A|B) = P(A)\)

 
Example  

The following is a data collected by a survey of 200 random people.

 Favorite Pet / Favorite Sport  Basketball  Football  Baseball  Sum
 Dog  47  54  10  111
 Cat  25  22  31  78
 Rabbit  2  3  6  11
 Sum  74  79  47  200

 

What is P(Rabbit)?

This is the probability that someone's favorite pet is a rabbit.

 Favorite Pet / Favorite Sport  Basketball  Football  Baseball  Sum
 Dog  47  54  10  111
 Cat  25  22  31  78
 Rabbit  2  3  6  11
 Sum  74  79  47  200

The answer is the sum of all the people who like rabbits (11), divided by the total (200).

P(Rabbit) = 11 / 200 = .055

What is P(Rabbitc)?

This is the probability that someone's favorite pet is not a rabbit.

 Favorite Pet / Favorite Sport  Basketball  Football  Baseball  Sum
 Dog  47  54  10  111
 Cat  25  22  31  78
 Rabbit  2  3  6  11
 Sum  74  79  47  200

The answer is the sum of all the people (200) minus the sum of all the people who like rabbits (11), divided by the total (200).

P(Rabbitc) = (200-11) / 200 = 189 / 200 = .945

What is P(Dog and Basketball)

This is the probability that someone's favorite pet is a dog, and that their favorite sport is basketball.

 Favorite Pet / Favorite Sport  Basketball  Football  Baseball  Sum
 Dog  47  54  10  111
 Cat  25  22  31  78
 Rabbit  2  3  6  11
 Sum  74  79  47  200

The answer is the amont of people who chose basketball and a dog (47), divided by the total (200).

P(Dog and Basketball) = 47 / 200 = .235

What is P(Football or Cat)?

This is the probability that someone chose football as their favorite sport, or a cat as their favorite pet.

 Favorite Pet / Favorite Sport  Basketball  Football  Baseball  Sum
 Dog  47  54  10  111
 Cat  25  22  31  78
 Rabbit  2  3  6  11
 Sum  74  79  47  200

Use the equation P(A or B) = P(A) + P(B) - P(A and B)

P(Football or Cat) = P(Football) + P(Cat) - P(Football and Cat)

P(Football or Cat) = (79 / 200) + (78 / 200) - (22 / 200) = .395 + .39 - .11 = .675

What is the probability of P(Cat and Dog)?

 Favorite Pet / Favorite Sport  Basketball  Football  Baseball  Sum
 Dog  47  54  10  111
 Cat  25  22  31  78
 Rabbit  2  3  6  11
 Sum  74  79  47  200

Since there is no overlap of Dog and Cat on the table, they are disjoint, or mutually exclusive.

If A and B are disjoint, P(A and B) = 0

P(Dog and Cat) = 0

What is the probability of P(Cat or Rabbit)?

 Favorite Pet / Favorite Sport  Basketball  Football  Baseball  Sum
 Dog  47  54  10  111
 Cat  25  22  31  78
 Rabbit  2  3  6  11
 Sum  74  79  47  200

Since there is no overlap of Rabbit and Cat on the table, they are disjoint, or mutually exclusive.

If A and B are disjoint, P(A or B) = P(A) + P(B)

P(Cat or Rabbit) = P(Cat) + P(Rabbit) 

P(Cat or Rabbit) = (78 / 200) + (11 / 200) = .39 + .055 = .445

What is P(Dog | Baseball)?

 Favorite Pet / Favorite Sport  Basketball  Football  Baseball  Sum
 Dog  47  54  10  111
 Cat  25  22  31  78
 Rabbit  2  3  6  11
 Sum  74  79  47  200

The rule is P(A | B) = P(A and B) / P(B)

P(Dog | Baseball) = P(Dog and Baseball) / P(Baseball)

P(Dog | Baseball) = (10 / 200) / (47 / 200) = (.05 / .235) = .213

What is the Probability of someone's favorite pet being a cat given that their favorite sport is basketball?

The rule is given that P(A | B) = Probability A given B

Applying it here the question is P(Cat | Basketball)

 Favorite Pet / Favorite Sport  Basketball  Football  Baseball  Sum
 Dog  47  54  10  111
 Cat  25  22  31  78
 Rabbit  2  3  6  11
 Sum  74  79  47  200

The rule is P(A | B) = P(A and B) / P(B)

P(Cat | Basketball) = P(Cat and Basketball) / P(Basketball)

P(Cat | Basketball) = (25 / 200) / (74 / 200) = (.125 / .37) = .338

What is the probability that of two random people, one's favorite sport is basketball and the other's is football?

 Favorite Pet / Favorite Sport  Basketball  Football  Baseball  Sum
 Dog  47  54  10  111
 Cat  25  22  31  78
 Rabbit  2  3  6  11
 Sum  74  79  47  200

These are two independent events as the probability of one person's favorite sport does not affect another's.

Independent events follow the rule: P(A and B) = P(A)•P(B) 

Applied here P(A and B) = P(Basketball and Football)

P(Basketball and Football) = P(Basketball)•P(Football)

P(Basketball and Football) = (74 / 200) • (79 / 200) = .37•.395 = .146