Probability - Event

Slide 1: Probability - Event

In probability, an event is a particular outcome or a group of outcomes of a random experiment.
Events can be classified as certain events, impossible events, favorable events, or unfavorable events.
We represent events by capital letters like A, B, C, etc.
The set of all possible outcomes of an experiment is known as the sample space, denoted by S.
An event is a subset of the sample space, denoted by A ⊆ S.

Slide 2: Sample Space

The sample space is the set of all possible outcomes of an experiment.
We represent the sample space by the symbol S.
The sample space can be finite, infinite, or empty.
For example, if we toss a fair coin, the sample space is {H, T}, where H represents heads and T represents tails.

Slide 3: Certain and Impossible Events

A certain event is an event that is guaranteed to occur, and its probability is 1.
An impossible event is an event that cannot occur, and its probability is 0.
For example, if we roll a fair six-sided die, the event of getting a number less than 7 is a certain event, while the event of getting a number greater than 10 is an impossible event.

Slide 4: Favorable and Unfavorable Events

A favorable event is an event that we desire to occur, and its probability lies between 0 and 1.
An unfavorable event is an event that we do not desire to occur, and its probability lies between 0 and 1.
For example, if we draw a card from a deck of 52 playing cards, the event of drawing a heart (one of the 13 hearts) is a favorable event, while the event of drawing a spade (one of the 13 spades) is an unfavorable event.

Slide 5: Intersection of Events

The intersection of two events A and B, denoted by A ∩ B, is the set of outcomes that belong to both A and B.
In simple terms, it represents the outcomes that satisfy both A and B.
For example, if A represents the event of rolling an even number on a fair six-sided die and B represents the event of rolling a number less than 4, then A ∩ B represents the event of rolling an even number less than 4, which is {2}.

Slide 6: Union of Events

The union of two events A and B, denoted by A ∪ B, is the set of outcomes that belong to either A or B or both.
In simple terms, it represents the outcomes that satisfy either A or B or both.
For example, if A represents the event of rolling an even number on a fair six-sided die and B represents the event of rolling a number less than 4, then A ∪ B represents the event of rolling an even number or a number less than 4, which is {1, 2, 3, 4, 6}.

Slide 7: Complementary Event

The complement of an event A, denoted by A’, is the set of outcomes in the sample space S that do not belong to A.
In simple terms, it represents the outcomes that do not satisfy A.
For example, if A represents the event of rolling an even number on a fair six-sided die, then A’ represents the event of rolling an odd number, which is {1, 3, 5}.

Slide 8: Mutually Exclusive Events

Two events A and B are said to be mutually exclusive if they cannot occur at the same time.
In other words, if A occurs, then B cannot occur, and vice versa.
The intersection of mutually exclusive events is always an empty set.
For example, if A represents the event of rolling a number less than 4 on a fair six-sided die and B represents the event of rolling a number greater than 4, then A and B are mutually exclusive.

Slide 9: Addition Law of Probability

The addition law of probability states that for any two events A and B, the probability of the union of A and B is equal to the sum of their individual probabilities minus the probability of their intersection.
Mathematically, it can be represented as P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
The addition law is valid for both mutually exclusive and non-mutually exclusive events.

Slide 10: Multiplication Law of Probability

The multiplication law of probability states that for any two events A and B, the probability of the intersection of A and B is equal to the product of their individual probabilities.
Mathematically, it can be represented as P(A ∩ B) = P(A) * P(B | A), where P(B | A) is the conditional probability of event B given that event A has already occurred.
The multiplication law is valid for both independent and dependent events. Sure! Here are the slides 11 to 20 on the topic of Probability - Event:

Slide 11: Independent Events

Two events A and B are said to be independent if the occurrence or non-occurrence of one event does not affect the occurrence or non-occurrence of the other event.
Mathematically, it can be represented as P(A ∩ B) = P(A) * P(B).
For example, if A represents the event of getting a head on a fair coin toss and B represents the event of rolling a number less than 4 on a fair six-sided die, then A and B are independent events.

Slide 12: Dependent Events

Two events A and B are said to be dependent if the occurrence or non-occurrence of one event affects the occurrence or non-occurrence of the other event.
Mathematically, it can be represented as P(A ∩ B) = P(A) * P(B | A), where P(B | A) is the conditional probability of event B given that event A has already occurred.
For example, if A represents the event of drawing a red card from a deck of playing cards and B represents the event of drawing a black card from the remaining cards after A, then A and B are dependent events.

Slide 13: Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already occurred.
It is denoted by P(B | A), which represents the probability of event B occurring given that event A has already occurred.
The formula for conditional probability is given by P(B | A) = P(A ∩ B) / P(A), where P(A ∩ B) is the probability of the intersection of events A and B and P(A) is the probability of event A.
For example, if A represents the event of drawing a king from a deck of playing cards and B represents the event of drawing a heart from the remaining cards after A, then P(B | A) represents the probability of drawing a heart given that a king has already been drawn.

Slide 14: Multiplication Rule for Independent Events

The multiplication rule for independent events can be used to find the probability of the intersection of independent events.
It states that for any two independent events A and B, the probability of the intersection of A and B is equal to the product of their individual probabilities.
Mathematically, it can be represented as P(A ∩ B) = P(A) * P(B).
For example, if A represents the event of rolling a number less than 4 on a fair six-sided die and B represents the event of flipping a head on a fair coin toss, then P(A ∩ B) represents the probability of rolling a number less than 4 and flipping a head.

Slide 15: Addition Rule for Mutually Exclusive Events

The addition rule for mutually exclusive events can be used to find the probability of the union of mutually exclusive events.
It states that for any two mutually exclusive events A and B, the probability of the union of A and B is equal to the sum of their individual probabilities.
Mathematically, it can be represented as P(A ∪ B) = P(A) + P(B).
For example, if A represents the event of drawing a spade from a deck of playing cards and B represents the event of drawing a heart from the same deck, then P(A ∪ B) represents the probability of drawing a spade or a heart.

Slide 16: Conditional Probability with Independent Events

If two events A and B are independent, the conditional probability of event B given that event A has already occurred is equal to the probability of event B.
Mathematically, it can be represented as P(B | A) = P(B).
For example, if A represents the event of rolling an even number on a fair six-sided die and B represents the event of flipping a head on a fair coin toss, then P(B | A) represents the probability of flipping a head given that an even number has already been rolled.

Slide 17: Bayes’ Theorem

Bayes’ theorem is a mathematical formula used to calculate the conditional probability of an event based on prior knowledge.
It is named after the Reverend Thomas Bayes.
Bayes’ theorem states that P(A | B) = P(B | A) * P(A) / P(B), where P(A | B) is the conditional probability of event A given that event B has occurred, P(B | A) is the conditional probability of event B given that event A has occurred, P(A) is the probability of event A, and P(B) is the probability of event B.
Bayes’ theorem is widely used in the fields of statistics, probability theory, and machine learning.

Slide 18: Permutations

Permutations are arrangements of objects or events in a particular order.
The number of permutations of n objects taken r at a time, denoted by nPr, can be calculated using the formula nPr = n! / (n - r)!, where n is the total number of objects and r is the number of objects taken at a time.
For example, if we have a deck of 52 playing cards and we want to arrange 5 cards in a particular order, the number of permutations is calculated as 52P5.

Slide 19: Combinations

Combinations are selections of objects or events without considering the order.
The number of combinations of n objects taken r at a time, denoted by nCr or C(n, r), can be calculated using the formula nCr = n! / (r!(n - r)!), where n is the total number of objects and r is the number of objects taken at a time.
For example, if we have a deck of 52 playing cards and we want to select 5 cards without considering the order, the number of combinations is calculated as 52C5.

Slide 20: Factorial

Factorial is a mathematical operation that denotes the product of all positive integers less than or equal to a given positive integer.
It is denoted by the symbol !.
For example, 5! (read as “5 factorial”) is calculated as 5 * 4 * 3 * 2 * 1, which is equal to 120.
Factorial is used in various mathematical calculations, such as permutations and combinations.

This is the end of slides 11 to 20 on the topic of Probability - Event. Sure! Here are slides 21 to 30 on the topic of Probability - Event:

Slide 21: Conditional Probability with Dependent Events

If two events A and B are dependent, the conditional probability of event B given that event A has already occurred is calculated using the formula P(B | A) = P(A ∩ B) / P(A), where P(A ∩ B) is the probability of the intersection of events A and B and P(A) is the probability of event A.
For example, if A represents the event of drawing a red card from a deck of playing cards and B represents the event of drawing a black card from the remaining cards after A, then P(B | A) represents the probability of drawing a black card given that a red card has already been drawn.

Slide 22: Probability of Independent Events

The probability of the intersection of independent events A and B is calculated using the formula P(A ∩ B) = P(A) * P(B).
For example, if A represents the event of rolling a number less than 4 on a fair six-sided die and B represents the event of flipping a head on a fair coin toss, then P(A ∩ B) represents the probability of rolling a number less than 4 and flipping a head.

Slide 23: Probability of Mutually Exclusive Events

The probability of the union of mutually exclusive events A and B is calculated using the formula P(A ∪ B) = P(A) + P(B).
For example, if A represents the event of drawing a spade from a deck of playing cards and B represents the event of drawing a heart from the same deck, then P(A ∪ B) represents the probability of drawing a spade or a heart.

Slide 24: Conditional Probability Examples

Example 1: A bag contains 5 red balls and 3 blue balls. If 2 balls are drawn without replacement, find the probability of drawing a red ball followed by a blue ball. Solution: Let A be the event of drawing a red ball and B be the event of drawing a blue ball. P(A) = 5/8 and P(B | A) = 3/7. Therefore, P(A ∩ B) = P(A) * P(B | A) = (5/8) * (3/7) = 15/56.
Example 2: A box contains 4 green marbles and 6 blue marbles. Two marbles are drawn at random. Find the probability of both marbles being green. Solution: Let A be the event of drawing a green marble and B be the event of drawing another green marble. P(A) = 4/10 and P(B | A) = 3/9. Therefore, P(A ∩ B) = P(A) * P(B | A) = (4/10) * (3/9) = 2/15.

Slide 25: Addition Rule Examples

Example 1: A fair six-sided die is rolled. Find the probability of rolling an even number or a number less than 4. Solution: Let A be the event of rolling an even number and B be the event of rolling a number less than 4. P(A) = 3/6, P(B) = 3/6, and P(A ∩ B) = 2/6. Therefore, P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = (3/6) + (3/6) - (2/6) = 4/6.
Example 2: A bag contains 5 red balls and 7 blue balls. Two balls are drawn at random. Find the probability of drawing at least one red ball. Solution: Let A be the event of drawing a red ball and B be the event of not drawing a red ball (drawing two blue balls). P(A) = 5/12 and P(B) = 7/12. Since A and B are mutually exclusive (cannot occur at the same time), P(A ∪ B) = P(A) + P(B) = (5/12) + (7/12) = 1.

Slide 26: Multiplication Rule Examples

Example 1: A box contains 3 yellow marbles and 4 green marbles. Two marbles are drawn at random without replacement. Find the probability of drawing two green marbles. Solution: Let A be the event of drawing a green marble and B be the event of drawing another green marble. P(A) = 4/7 and P(B | A) = 3/6. Therefore, P(A ∩ B) = P(A) * P(B | A) = (4/7) * (3/6) = 2/7.
Example 2: A deck of 52 playing cards is shuffled. Two cards are drawn at random without replacement. Find the probability of drawing a king and then a queen. Solution: Let A be the event of drawing a king and B be the event of drawing a queen. P(A) = 4/52 and P(B | A) = 4/51. Therefore, P(A ∩ B) = P(A) * P(B | A) = (4/52) * (4/51) = 16/2652.

Slide 27: Permutations Examples

Example 1: How many different 3-letter words can be formed using the letters A, B, C, D, E, and F without repetition? Solution: The number of permutations is calculated as 6P3 = 6! / (6 - 3)! = 6! / 3! = 6 * 5 * 4 = 120.
Example 2: In how many ways can 5 people be arranged in a line? Solution: The number of permutations is calculated as 5P5 = 5! / (5 - 5)! = 5! / 0! = 5 * 4 * 3 * 2 * 1 = 120.

Slide 28: Combinations Examples

Example 1: How many different 2-card hands can be dealt from a standard deck of 52 playing cards? Solution: The number of combinations is calculated as 52C2 = 52! / (2! * (52 - 2)!) = 52! / (2! * 50!) = (52 * 51) / (2 * 1) = 1326.
Example 2: In how many ways can 3 people be selected from a group of 10 people? Solution: The number of combinations is calculated as 10C3 = 10! / (3! * (10 - 3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120.

Slide 29: Factorial Examples

Example 1: Calculate 4!. Solution: 4! = 4 * 3 * 2 * 1 = 24.
Example 2: Calculate 6!. Solution: 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720.