Probability and Counting Principles

College Algebra

Counting Principles

According to the Addition Principle, if one event can occur in 𝑚 ways and a second event with no common outcomes can occur in 𝑛 ways, then the first or second event can occur in 𝑚 + 𝑛 ways.

According to the Multiplication Principle, if one event can occur in 𝑚 ways and a second event can occur in 𝑛 ways after the first event has occurred, then the two events can occur in 𝑚×𝑛 ways. This is also known as the Fundamental Counting Principle.

Example: On a restaurant menu, there are 3 appetizer options, 2 vegetarian entrée options and 5 meat entrée options, and 2 dessert options.There are 3× 2 + 5 ×2 = 42 different choices for a three-course dinner.

Permutations of Distinct Objects

Given 𝑛 distinct objects, the number of ways to select 𝑟 objects from the set in order is:

𝑃 𝑛, 𝑟 =𝑛!

𝑛 − 𝑟 !

Example: A club with six people need to elect a president, a vice president and a treasurer. How many ways can the officers be elected?

Solution: 𝑃 6,3 = 0!012 !

= 0343532363723637

= 6 3 5 3 4 = 120

Permutations of Non-Distinct Objects

If there are 𝑛 elements in a set and 𝑟7 are alike, 𝑟6 are alike, 𝑟2 are alike, and so on through 𝑟:, the number of permutations can be found by:

𝑛!𝑟7! 𝑟6!⋯ 𝑟:!

Example: Find the number of rearrangements of the letters in the word DISTINCT.Solution: There are 8 letters in the word, but I and T are repeated two times each. Therefore, the number of permutations is:

8!2! 3 2! =

8 3 7 3 6 3 5 3 4 3 3 3 2 3 12 3 2 = 10080

Combinations

When we are selecting objects and the order does not matter, we are dealing with combinations.Given 𝑛 distinct objects, the number of ways to select 𝑟 objects from the set is:

𝐶 𝑛, 𝑟 =𝑛!

𝑟! 𝑛 − 𝑟 !

Example: An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split?

Solution: 𝐶 10,3 = 7?!2! 7?12 !

= 7?!2!@!

= 7?3A3B3@!2363@!

= @6?0= 120

Number of Subsets of a Set

A set containing 𝑛 distinct objects has 2C subsets.

Example: A restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a baked potato?

Solution: There are 4 options, so there are 25 = 16 possible ways to order a baked potato.This result is the same as:

𝐶 4,0 + 𝐶 4,1 + 𝐶 4,2 + 𝐶 4,3 + 𝐶 4,4 = 1 + 4 + 6 + 4 + 1 = 16

Binomial Coefficients

In the shortcut to finding 𝑥 + 𝑦 C we use combinations to find the coefficients that will appear in the expansion of the binomial.If 𝑛 and 𝑟 are positive integers with 𝑛 ≥ 𝑟, then the binomial coefficientis:

𝑛𝑟 = 𝐶 𝑛, 𝑟 =

𝑛!𝑟! 𝑛 − 𝑟 !

Note that CG = C

C1G .

Example: A6 = A!

6! A16 != A!

6!@!= A3B3@!

63@!= 36

A@ = A!

@! A1@ != A!

@!6!= A3B3@!

@!36= 36

Binomial Theorem

The Binomial Theorem is a formula that can be used to expand any binomial.

𝑥 + 𝑦 C = ∑ C: 𝑥

C1:𝑦:C:I?

= 𝑥C + C7 𝑥

C17𝑦 + C6 𝑥

C17𝑦6 + ⋯+ CC17 𝑥𝑦

C17 + 𝑦C

Example: Expand 𝑥 + 𝑦 4

=50 𝑥4𝑦? +

51 𝑥5𝑦7 +

52 𝑥2𝑦6 +

53 𝑥6𝑦2 +

54 𝑥7𝑦5 +

55 𝑥?𝑦4

= 𝑥4 + 5𝑥5𝑦 + 10𝑥2𝑦6 + 10𝑥6𝑦2 + 5𝑥𝑦5 + 𝑦4

Using the Binomial Theorem to Find a Single Term

The (𝑟 + 1)th term of the binomial expansion of (𝑥 + 𝑦)C is:𝑛𝑟 𝑥C1G𝑦G

Example: Find the sixth term of 3𝑥 − 𝑦 A without fully expanding the binomial.Solution:Let 𝑟 = 5 for the sixth term, and use 3𝑥 and −𝑦 for the two variables.

95 3𝑥 A14 −𝑦 4 =

9 3 8 3 7 3 6 3 5!4! 3 5! 35𝑥5(−1)4𝑦4 = −10206𝑥5𝑦4

Probabilities

The likelihood of an event is known as probability. The probability of an event 𝑝 is a number that always satisfies 0 ≤ 𝑝 ≤ 1, where 0 indicates an impossible event and 1 indicates a certain event.

A probability model is a mathematical description of an experiment listing all possible outcomes and their associated probabilities.

The probability of an event 𝐸 in an experiment with sample space 𝑆 with equally likely outcomes is given by

𝑃 𝐸 =numberofelementsin𝐸numberofelementsin𝑆 =

𝑛(𝐸)𝑛(𝑆)

𝐸 is a subset of 𝑆, so it is always true that 0 ≤ 𝑃(𝐸) ≤ 1.

Probability for Multiple Events

The probability of the union of two events 𝐸 and 𝐹 (written 𝐸 ∪ 𝐹) equals the sum of the probability of 𝐸 and the probability of 𝐹 minus the probability of 𝐸 and 𝐹 occurring together (which is called the intersection of 𝐸and 𝐹 and is written as 𝐸 ∩ 𝐹).

𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹 − 𝑃(𝐸 ∩ 𝐹)

Example: A card is drawn from a standard deck. Find the probability of drawing a heart or a 7.

Solution: 𝑃 ℎ = 7246

, 𝑃 7 = 546

, and 𝑃 ℎ ∩ 7 = 746

𝑃 ℎ ∪ 7 = 7246+ 546− 746= 70

46= 5

72

Computing the Probability of Mutually Exclusive Events

The probability of the union of two mutually exclusive events 𝐸 and 𝐹 is given by

𝑃 𝐸 ∪ 𝐹 = 𝑃 𝐸 + 𝑃 𝐹

Example: A card is drawn from a standard deck. Find the probability of drawing a heart or a spade.

The events “drawing a heart” and “drawing a spade” are mutually exclusive because they cannot occur at the same time. The probability of drawing a heart is 7

5, and the probability of drawing a spade is also 7

5, so the probability

of drawing a heart or a spade is 75+ 75= 7

6

Probability That an Event Will Not Happen

The complement of an event 𝐸, denoted 𝐸′, is the set of outcomes in the sample space that are not in 𝐸.

𝑃 𝐸c = 1 − 𝑃(𝐸)

Example: Two six-sided dice are rolled. What is the probability that the sum of the numbers is greater than 3?Solution:The sample space is the set of all 36 possible outcomes from 1 + 1 to 6 + 6. It is easier to consider the 3 possible totals not greater than 3: 1 + 1, 1 + 2, and 2 + 1. Therefore, if 𝑃 𝐸c = 2

20= 7

76, then 𝑃 𝐸 = 77

76.

Computing Probability Using Counting TheoryMany probability problems use permutations and combinations to find the number of elements in events and sample spaces.

Example: A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears. Find the probability that 2 bears and 3 dogs are chosen.

Solution:There are 6 bears, so there are 𝐶(6,2) ways to choose 2 bears.There are 5 dogs, so there are 𝐶(5,3) ways to choose 3 dogs.There are 𝐶(6,2) 3 𝐶(5,3) ways to choose 2 bears and 3 dogs.There are 14 toys, so there are 𝐶 14,5 ways to choose 5 toys.

𝑃 𝐸 =𝐶(6,2) 3 𝐶(5,3)

𝐶(14,5) =15 3 102002 =

751001

Quick Review

• What is the Fundamental Counting Principle?• What is the difference between a permutation and a combination?• What is the formula for the number of permutations of non-distinct items?• How many subsets are there for a set of 𝑛 distinct items?• What is the formula for a binomial coefficient?• How is the Binomial Theorem used?• What is the sum of probabilities for all possible events in a given

probability model?• How do you compute the probability of the union of two events?