Probability: Experiments and Events

• Chapter 3 focuses on probability and its basic concepts.

Probability Experiments

• A probability experiment involves actions with specific results like counts, measurements, or responses.
• Rolling a die and noting the number rolled exemplifies a probability experiment.
• The outcome is the result of a single trial in a probability experiment.
• The sample space encompasses all possible outcomes of an experiment.
• The sample space when rolling a die has six outcomes: {1, 2, 3, 4, 5, 6}.

Events

• An event includes one or more outcomes and represents a subset of the sample space.
• Events are represented by uppercase letters.
• A simple event is an event with a single outcome.
• For example, rolling a die where event A is rolling an even number ({2, 4, 6}) is not a simple event.

Classical Probability

• Classical (or theoretical) probability applies when each outcome in the sample space is equally likely.
• The classical probability for an event E is calculated as the number of outcomes in event E divided by the total number of outcomes in the sample space.
• A die is rolled to find the probability of Event A: rolling a 5.
• A(Probability of Event A) = 1/6 ≈ 0.167 because there is one outcome in event A: {5}.

Empirical Probability

• Empirical (or statistical) probability is based on observations from probability experiments.
• The empirical frequency of an event E is its relative frequency.
• The probability of event E, P(E), equals the Frequency of Event E divided by the Total frequency, or f/n.
• A travel agent knows that 12 out of every 50 reservations are for a cruise
• The probability that the next reservation will be for a cruise equals P(cruise) = 12/50 = 0.24.

Law of Large Numbers

• As an experiment is repeated, the empirical probability of an event approaches the event's theoretical probability.
• Sally gets 3 heads, when flipping a coin 20 times
• An empirical probability of 3/20 is not representative of the theoretical probability of 1/2.
• With an increase in the number of coin tosses, the empirical probability gets closer to the theoretical probability (law of large numbers).

Probabilities with Frequency Distributions

• Given a frequency distribution of ages of students in a statistics class, the probability of a student's age falling within a range can be calculated.
• For a class of 30 students, with 8 students aged between 26 and 33, P (age 26 to 33) = 8/30 ≈ 0.267.

Subjective Probability

• Subjective probability is based on intuition, educated guesses, and estimates.
• For instance, a business analyst may estimate a 0.15 probability of a certain union going on strike.

Range of Probabilities Rule

• The probability of any event E is between 0 and 1, inclusive.
• This is written as 0 ≤ P(A) ≤ 1.
• An impossible event to occur is 0 , a certain event is a 1, a even chance is 0.5.

Complementary Events

• The complement of an event E consists of all outcomes in the sample space not included in E and is denoted as E' (read "E prime").
• P(E) + P(E') = 1, P(E) = 1 − P(E'), and P(E') = 1 – P(E)
• Calculating that the change of not drawing a blue chip when there are 5 red, 4 blue, and 6 white chips in a basket, P (selecting a blue chip) = 4/15 ≈ 0.267 and P (not selecting a blue chip) = 1 - 4/15 = 11/15 ≈ 0.733.

Conditional Probability

• A conditional probability is the likelihood of an event occurring, given that another event has already occurred, denoted as P(B|A), or "Probability of B, given A".
• With 5 red, 4 blue, and 6 white chips, the probability of selecting a red chip after selecting a blue chip (without replacement) equals P (selecting a red chip|first chip is blue) = 5/14 ≈ 0.357.

Conditional Probability

• Based on a survey of college students and hours of study time, the probability that a student studies more than 10 hours given they are male shows P (more than 10 hours|male) = 16/49 ≈ 0.327, out of 49 male students, 16 study for more than 10 hours per week.

Independent Events

• Two events are independent when the occurrence of one does not influence the probability of the other.
• Events A and B are independent if P(B|A) = P(B) or P (A|B) = P(A).
• Events not meeting this criteria are dependent.
• Selecting a diamond from a standard deck, replacing it, and then selecting a spade from the deck, example of independent events P(B|A) = 13/52 = 1/4 and P(B) = 13/52 = 1/4 and the probability are independent.

Multiplication Rule

• The probability of both events A and B occurring in sequence equals P(A and B) = P(A) • P(B|A).
• If events A and B are independent, then P (A and B) = P(A) • P (B).
• Selecting two cards without replacement determines dependent events as P (diamond and spade) = P (diamond) · P (spade |diamond) = 13/52 * 13/51 = 169/2652 ≈ 0.064.

Multiplication Rule

• Rolling a die and tossing 2 coins is an example where P (rolling a 5) = 1/6.
• Whether or not the roll is a 5, P (Tail ) =1/ 2 and so the events are independent.
• With an equal probability equals
• P (5 and T and T ) = P (5) P(T). P(T) = 1/6 * 1/2 * 1/2 = 1/24 ≈ 0.042.

Mutually Exclusive Events

• Two events are mutually exclusive if they cannot occur simultaneously.

Mutually Exclusive Events

• Rolling a number less than 3 on a die and rolling a 4 on a die is two Mutually exclusive events.

Mutually Exclusive Events

• Selecting a Jack from a deck of cards and Selecting a heart from a deck of cards is a no mutually exclusive events.

The Addition Rule

• The probability of occurrence of event A or B equals P(A or B) = P(A) + P(B) – P(A and B).
• For mutually exclusive events A and B, the rule simplifies to P (A or B) = P(A) + P(B).
• When rolling a die, the probability of rolling a number less than 3 or a 4, equals P (roll a number less than 3 or roll a 4) = P (number is less than 3) + P (4) = 2/6+ 1/6 = 3/6 equals 0.5.

The Addition Rule

• When drawing a card and the event is drawing A Jack or drawing a heart then the probability equals
• P (select a Jack or select a heart) = P (Jack) + P (heart) – P (Jack of hearts) = 4/52 + 13/52 – 1/52 = 16/52 ≈ 0.308.

The Addition Rule

• Out of 100 students, calculate those that spend between 5 and 10 hours or more than 10 hours with - P (5 to 10 hours or more than 10 hours) = P(5 to10) + P(10) = 46/100 + 30/100 = 76/100 = 0.76.

Fundamental Counting Principle

• If one event results in m ways and a second event shows as n ways, the two events happens in m. n ways.
• The rule can be extended for events happening in a sequence.
• In the meal selection calculation, total ways of selecting dishes in 4.2. 5 = 40.

Fundamental Counting Principle

• With 2 coins flipped, there are 2 × 2 = 4 different outcomes: {HH, HT, TH, TT}.

Fundamental Counting Principle

• Each digit is chosen from 0-9 to construct an access code for house. The access code consists of 5 digits.
• A total of 10.10.10.10. 10 = 100,000 codes happen when each digit can be repeated.
• Because each digit is only used once, there are 10.9.8.7.6 = 30,240 cases.

Permutations

• A permutation represents an ordered arrangement of objects.
• n distinct objects show n! unique permutations.

Permutations

• Calculating that the number of possible question arrangements in 7 questions is 7! = 7.6.5.4.3.2.1 =5040 surveys.

rPermutation of n Objects Taken at a Time

• Considering that you read 5 books from a list of 8, different order is calculated as nP₁ = 8P5 = 8! (8-5)! = 8.7.6.5.4.3.2.x 3.2.X = 6720 ways

Distinguishable Permutations

• The number of distinguishable permutations of n objects, where n₁ are one type, n₂ are another type, and so on is n! / n₁! ·n₂! ·n₃!✉ nk!.'
• Where the n₁ + n₂ + n₃ + + nk = n.
• Jessie is planing 10 plants and that will be the number of possible arrangements of the rose bushes (3), daffodils (4), and lilies (3) equals to 10!/3!4!3! = 10.9.8.7.6.5.4!/3!4!3! = 4,200

Combination of n Objects Taken at a Time

• A combination is selecting objects where order does not matter.
• The amount of way r objects are from a group of n objects is nCr = n!/(n-r)!r!.
• You are required to read 5 books from a list of 8 with no order then it is 8C5 =8.7.6.5! / 3!5!= 56 combinations

Application of Counting Principles

• Drawing 6 numbers (in any order) is the number of winning combinations when you draw from 44 numbers in a lottery.
• If there is only one prize then , 44C6 = 44!/(6!38!), and the Probability (Wining) = 1 out of 7,059,052 combination ≈ 0.00000014