Probability and Combinatorics: Key Concepts and Examples

Probability and Combinatorics: Key Concepts

Combinatorics provides procedures and formulas necessary to count the possibilities to choose a set of items with certain characteristics.

Permutations and Variations

  • By taking all elements of a finite set and ordering them in all possible ways, we have a permutation.
  • A variation of order k (or k-order variation) is a group of k elements chosen from a total set, where each group is different either by the elements it contains or by the order in which they are arranged.

Sample Space, Events, and Experiments

  • Sample space is the set of all possible outcomes of a process of observation or experiment.
  • An event is a subset of the sample space. Events are elements of the sample space that meet a given condition.
  • Events are denoted with capital letters, e.g., A, B, C, etc.

Probability of an Event

  • Events, denoted by letters A, B, C, etc., have a probability associated with each of them, designated as P(A), P(B), P(C), etc.
  • The probability of an event, from a classical viewpoint, is the ratio of the number of results that are favorable to the event and the total number of elementary and equally likely outcomes of the experiment.

Independent and Conditional Events

  • Independent Events: Two events are independent when the occurrence of one event has no effect on the occurrence of the other. The probability of one event does not affect the probability of the other.
  • Conditional Probability: Conditional probability applies when we already know the result of one of two events.

Examples and Exercises

Here are some examples to illustrate these concepts:

Example 1

In how many distinct ways can you arrange 3 oranges and 2 apples so that there are 2 fruits together?

Example 2

In how many ways can you arrange the 4 aces of a Spanish deck?

Example 3

If you have a box with 5 red balls, the event “drawing two red balls” is a possible outcome.

Example 4

Someone draws a card from a deck of 50 cards and simultaneously rolls a 6-sided die. How many elements are in the sample space?

Example 5

Someone draws a card from a Spanish deck. What is the probability of rolling a 7 on a die?

Example 6

For any event A, what is the interval in which its probability always lies?

Example 7

A number is chosen at random from 1 to 10. Calculate the probability of choosing an even number.

Example 8

Two dice numbered from 1 to 6 are thrown. Calculate the probability that the sum of the two dice is 11.

Example 9

A box contains 18 red balls and 12 white balls. What is the probability of drawing a white ball?

Example 10

A bag contains 18 glass marbles, 12 stone marbles, and 10 clay marbles. If a marble is drawn at random, what is the probability that it is glass or stone?

Example 11: Expected Value

A raffle ticket offers two prizes: $50,000 with a probability of 0.001 and $20,000 with a probability of 0.003. What would be a fair price to pay for a ticket, based on mathematical expectation?

Solution: The expected value is ($50,000)(0.001) + ($20,000)(0.003) = $50 + $60 = $110, which is a fair price.

Example 12: Expected Value in a Risky Business

In a risky business, a woman can earn $3,000 with a probability of 0.6 or lose $1,000 with a probability of 0.4. Find the expected value.

Solution: The expected value is ($3,000)(0.6) + (-$1,000)(0.4) = $1,800 – $400 = $1,400.

Example 13: Expected Value with Multiple Participants

A bag contains 2 white balls and 3 black balls. Four people, A, B, C, and D, in that order, each draw a ball without replacement. The first to draw a white ball receives $10. Determine the expected value for A, B, C, and D.

Solution:

  • P(A) = P(A wins) = 2 / (3+2) = 2/5. The expected value of A = (2/5)($10) = $4.
  • P(A loses and B wins) = P(A)P(B/A) = (3/5)(2/4) = 3/10. The expected value of B = $3.
  • P(A and B lose, C wins) = P(A)P(B/A)P(C/AB) = (3/5)(2/4)(2/3) = 1/5. The expected value of C = $2.
  • P(A, B, and C lose, D wins) = P(A)P(B/A)P(C/AB)P(D/ABC) = (3/5)(2/4)(1/3)(1/1) = 1/10. The expected value of D = $1.

Check: $4 + $3 + $2 + $1 = $10 and 2/5 + 3/10 + 1/5 + 1/10 = 1

Example 14: Forming Different Groups

How many different bouquets can be formed with 5 different varieties of flowers?

Solution: Each flower can either be chosen or not. These two possibilities occur for each flower, resulting in a total of 2^5 possibilities. However, we must exclude the case where no flower is chosen. Therefore, the number of bouquets is 2^5 – 1 = 31.

Alternatively: (5 choose 1) + (5 choose 2) + (5 choose 3) + (5 choose 4) + (5 choose 5) = 5 + 10 + 10 + 5 + 1 = 31.

In general, for any positive integer n: (n choose 1) + (n choose 2) + … + (n choose n) = 2^n – 1.

Example 15: Forming Words

With 7 consonants and 5 vowels, how many words can be formed that have 4 different consonants and 3 different vowels? Meaningless words are allowed.

Solution: The 4 consonants can be chosen in (7 choose 4) ways. The 3 vowels can be chosen in (5 choose 3) ways. The 7 chosen letters can be arranged in 7! ways. So, the required number is (7 choose 4)(5 choose 3)7! = 35 * 10 * 5040 = 1,764,000.

Example 16: Drawing Balls from a Box

A box contains 8 red balls, 3 white balls, and 9 blue balls. If 3 balls are drawn at random, determine the probability that:

a) All 3 are red.

b) 2 are red and 1 is white.

Solution:

a) Let R1, R2, and R3 denote the events that the first, second, and third balls drawn are red, respectively. Then, P(R1 and R2 and R3) = P(R1)P(R2/R1)P(R3/R1 and R2) = (8/20)(7/19)(6/18) = 14/285.

b) P(2 are red and 1 is white) = [(8 choose 2)(3 choose 1)] / (20 choose 3) = 7/95.

Statistical Examples

Example 17: Physical Examination Scores

In a physical examination, the ideal score is 150 points, and the standard deviation is 18.

a) Calculate the mean.

b) Determine the standard unit scores for students who obtained: i) 35 ii) 60 iii) 100

c) Find the scores corresponding to standard scores of: i) -2 ii) 2.95

Solution:

a) 3.5 = (150 – x)/18 => 63 = 150 – x => x = 150 – 63 => x = 87

b) Z = (35 – 87)/18 = -2.89; Z = (60 – 87)/18 = -1.5; Z = (100 – 87)/18 = 0.72

c) -2 = (x – 87)/18 => x = 51; 2.95 = (x – 87)/18 => x = 140.1

Example 18: Calculating Areas Under the Standard Normal Curve

Find the area under the standard normal curve:

a) Between Z = 0 and Z = 1.7

b) Between Z = 0 and Z = -0.68

c) Between Z = -0.56 and Z = 2.41

d) Between Z = 0.91 and Z = 2.94

e) To the left of Z = -0.9

f) To the right of Z = -1.28

g) To the right of Z = 3.05 and to the left of Z = -1.54

Solution:

a) The area is 0.4554

b) P(-0.68 < Z < 0) = 0.2518

c) P(-0.56 < Z < 0) = 0.2123, P(0 < Z < 2.41) = 0.4920; Total = 0.2123 + 0.4920 = 0.7043

d) P(0.91 < Z < 2.94) = P(0 < Z < 2.94) – P(0 < Z < 0.91) = 0.4984 – 0.3186 = 0.1798

e) 0.5 – P(-0.9 < Z < 0) = 0.5 – 0.3159 = 0.1841

f) P(-1.28 < Z < 0) + 0.5 = 0.3997 + 0.5 = 0.8997

g) P(0 < Z < 3.05) = 0.4989; To the right: 0.5 – 0.4989 = 0.0011

P(-1.54 < Z < 0) = 0.4382; To the left: 0.5 – 0.4382 = 0.0618

Total = 0.0011 + 0.0618 = 0.0629

Example 19: Student Weights

The average weight of 1000 male students from a certain college is 62 kilograms (kg), and the standard deviation is 8 kg. Assume that the weights are normally distributed. Determine how many students weigh:

a) Between 54.4 and 70 kg

b) More than 80 kg

c) Less than 47.4 kg

Solution:

a) 54.4 units = (54.4 – 62)/8 = -0.95; 70 units = (70 – 62)/8 = 1

P(-0.95 < Z < 0) + P(0 < Z < 1) = 0.3289 + 0.3413 = 0.6702; 0.6702 * 1000 = 670.2 or approximately 670 students.

b) More than 80 kg: (80 – 62)/8 = 2.25; 0.5 – 0.4878 = 0.0122; 0.0122 * 1000 = 12.2 or approximately 12 students.

c) Less than 47.4 kg: (47.4 – 62)/8 = -1.825; 0.5 – 0.4664 = 0.0336; 0.0336 * 1000 = 33.6 or approximately 34 students.