And the probability that the student will pass the final

A subset of a set is a sub-collection of the set. For example, if S = {1, 4, 10}, then A = {1, 4} is a subset of S since A is a sub-collection of S. (We say that every element of A is an element of S.) If A is a subset of S, we write A ⊆ S. By definition, every set is a subset of itself. If A is a subset of S but A ̸= S, we say that A is a proper subset of S. If A is a proper subset of S, we write A ⊂ S.

The set that contains no object is the empty set, and is denoted by the math-ematical symbol ∅. The empty set is a subset of any set.

If A, B are finite sets, then:

|A ∪ B| = |A| + |B| − |A ∩ B|
Notice that |∅| = 0
Definition: We say that two sets A and B are mutually exclusive if A∩B = ∅Formula: If A and B are mutually exclusive, then:

The multiplication principle can be expanded to cases in which more than 2 experiments are being performed. If experiment 1 can be performed in m1 many ways, and experiment 2 can be performed in m2 many ways, ..., and experiment k can be performed in mk many ways, then there are m1m2 · · · mk many different ways in which all the combined experiments may be performed.

Example: In how many different ways can one buy a phone, a radio, and a television of each if there are 10 different brands of phone, 6 different brands of radio, and 8 different brands of TV to choose from?

Example: How many different words can we form by using the letters of the word MONEY? Note: By a word here we do not mean a word that is necessarily in the dictionary, just any combination of the alphabets. Examples of ”words”would be MNOEY, MEOYN, OENMY ...etc.

Notice that all five of the letters in the word money are different. We may use any of the five letters to be the first letter of the word, so for the first letter of the word we have 5 choices. On the second letter of the word we have 4 choices left, on the third letter we have 3 choices left, then we have 2 choices for the fourth letter, then only 1 choice for the last letter. Using the counting principle we have 5 · 4 · 3 · 2 · 1 = 5! = 120 different words that can be formed.

Notice in this example that there is no distinction between any of the students in the group, as long as the student is in the group, he/she is a class representative. This means that the order in which we choose the group of 6 does not matter.

If we have n many objects, and we choose r from the group, and the order does not matter, then each of such choice is called a combination.

Sometimes, we may not have all the objects in a collection to be distinct, but instead represent two or more kinds of the same object. In this case care needs to be taken when we want to find the number of combinations or permutations.

Example:

Example:

There are 11 kinds of donuts, 7 kinds of bagels, and 9 different favors of soda. If you want to buy 6 donuts, 3 bagels, and 3 sodas, how many different ways can you buy?

Example:

Three biologists, four economists, and seven mathematicians are to sit in a row to watch a movie. If those of the same profession insist to be seated together, how many different ways can these people be seated?

11!

4! · 4! · 2!= 34650
In general, if there are n many objects to be arranged, but among these objects, n1 of one kind is identical, and n2 of another kind is identical, n3 of third kind is identical ..., then the number of different permutations is given by:

Definition: An experiment is a procedure that produces an outcome. For ex-ample, the toss of a coin is an experiment (and the outcome would be head or tail). The rolling of a die is an experiment, and the outcome is 1, 2, 3, 4, 5, or 6.

The sample space of an experiment is the set that contains all possible out-comes of the experiment

Intuitively, the number P({e}) is the probability that a simple event (one partic-ular outcome) {e} will happen when the experiment is performed. The closer to 1, the more likely the event will happen. The closer to 0, the less likely the event will happen. If P({e}) = 1, then event {e} will always happen if the experiment is performed, if P({e}) = 0, event {e} will never happen.

b. The sum of the probabilites over all the simple events of the sample space is one. That is, if {e1}, {e2}, {e3}, ..., {en} are all the simple events of the sample space, then

P(A) =|A|
n
where |A| is the order of A (number of elements in A).

That is, the probability of an event happening in which every outcome is equally likely, is equal to the number of outcomes of the event divided by the total number of possible outcomes.

If a fair die is rolled, what is the probability that the number is less than 7? The event A here is number less than 7, this is the set {1, 2, 3, 4, 5, 6}. This set has 6 elements. The sample space also has 6 elements. The probability of event A is P(A) = 6 6 = 1. This means that this event is destined to happen on any try of the experiment.

In calculating probability using the above formula, it is important that each outcome in the sample space is equally likely. We will use the terms like fair die, fair coin, or choosing a number in random to mean that each outcome of the experiment is equally likely.

|A ∪ B| = |A| + |B| − |A ∩ B|,
we have an equivalent version in probability:

Suppose A and B are two events, then

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

Law of the Complement:

Ans: There are 8 letters in the word, so the size of the sample space S is 8, the desirable event is vowel, which is the set {a, o, e}, which has a size of 3, using the probability formula, the probility of this is:

3

{(5, 5), (4, 6), (6, 4), (5, 6), (6, 5), (6, 6)} , so the probability of the event is:

36 = 1

ways two cards can be dealt from a 52 card deck. There’s only 4 Aces, and we�52�= 1326 different

need to choose two from them, so there’s�4�= 6 ways two Aces can be dealt.

A group of lawyers consist of 5 women and 7 men. If 4 lawyers are chosen

randomly to represent a client, what is the probability that all of them will be

ways to do so. There is 495≈ 0.010 ≈ 1% chance.

Example: The California lottary requires you to pick 5 numbers from the numbers

5-number (from 56) and 1-number (from 46) combinations are possible. For this

problem, the order in which you pick the numbers do not matter, so this is a

size of the event (the numbers of your ticket) is 1, so your chance of winning is:

1
175, 711, 536.

provided that P(B) ̸= 0.

This formula says that, the probability of A happening given that B has already happened is given by the probability of A and B divided by the probability of B.

Let B be the event of ”person is a man”.

Example: The probability that a student will pass the midterm is 60%, and the probability that the student will pass the final is 53%. The probability that the student will pass both tests is 45%. If a student has already passed the midterm, what is the probability the she will pass the final exam?

Let A be the event of ”passing the final”

Mathematically, we can see that if two events A and B are independent, then the probability of one given that the other has occured should not change. That is,

Definition: Two events A and B are independent events if P(A|B) = P(A) and P(B|A) = P(B).

That is, if two events are independent, then the probability that they both happen is just the product of the probability of them happening individually.

Example:

Birthday Paradox In a room of 40 people, what is the probability that at least two people will have the same birthday?

It is not easy to find this probability directly. Instead, we first calculate the probability that no two people will have the same birthday.

The probability that no two people would have the same birthday is about 11%. Using the law of the complement, this means the probability that at least two people would have the same birthday is 1 − 0.109 ≈ 89%.