Probability of independent/dependent events
When two events are independent, the fact that A occurs does not affect the probability of B occurring.
You need to use this formula:
For example:
In a bag of marbles, there are 3 red and 5 blues.
If two marbles are chosen from the jar with replacement.
What is the probability of the two marbles being red?
P(red) = 3/(3+5) = 3/8
Since you are choosing the marble and then putting it back in the bag, the probability is INDEPENDENT
P(red and red) = (3/8)(3/8) = 9/64
When two events are dependent, the probability of A affects the probability of B
When two events are dependent use this formula:
If I have the same problem but WITHOUT REPLACEMENT
In a bag of marbles, there are 3 red and 5 blues.
If two marbles are chosen from the jar without replacement.
What is the probability of the two marbles being red?
In this case the probability is DEPENDENT, that is once the red marble is chosen the first time P(red): 3/8, the second time there are only 2 out 3 red marbles, and the bag only has 7 marbles, therefore the probability of choosing two red marbles is:
P(red and red) = (3/8)(2/7) = 3/28
Probability of Mutually Exclusive Events (Disjoint)
Two events are mutually exclusive when
they can NOT happen at the same time.
For example:
Getting heads or tails, being female or male, being democrat or republican.
Probability of Non- Mutually
Exclusive Events (Joint)
When two events are non- mutually exclusive,
they can happen at the same time,
We can use this formula:
For example:
The probability of being a woman and being democrat, the probability of being a sport player and being a male.
Let's say you have two events, P(A) and P(B).
If P(A)=0.60 and P(B)=0.30 are independent and not mutually exclusive, then find P(A or B)
1) Find P(A and B) first:
P(A and B) = (0.6)(0.30) = 0.18
2) Now apply the formula:
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = 0.60 + 0.30 - 0.18 = 0.72
Factorial Notation
The factorial of N! is the product of all positive integers from N to 1.
For example:
We can do operations with factorials,
For example:
5! - 4! = (1•2•3•4•5)-( 1•2•3•4)= 120 - 24 = 96
6!/5!= (1•2•3•4•5•6)/(1•2•3•4•5) = 6
Counting Methods
If you have “m” ways of doing one thing and then “n” ways of doing another then m*n is how many ways you can do both, this sentence represents the fundamental principle of counting.
For example:
Alex can choose between 4 appetizers, 5 main meals, and 3 desserts. How many different meals can he order?
Counting: 4*5*3= 60 meals
Combination and Permutation
When you have to select items from a collection, you need to do a combination or a permutation, depending on if the order matters.
For example: you have 4 ice cream flavors:
Vanilla (V), Chocolate (C), Orange (O), Mint (M).
How many ways can you chose 2 flavors out of the 4?
You can choose the following ways:
VC, VO, VM, CO, CM, OM, that is 6 ways.
Does it matter if you eat vanilla first and then chocolate?
NO, because VC is the same as CV, the order DOES NOT MATTER, then you have to use a combination.
Using the formula
n=4
r=2
Let’s do a permutation example:
You have 6 candidates running for the board of an association.
The board requires a President (P), Vice-president (VP) and a Treasurer (T). How many ways can you select the board?
There are 6 candidates and 3 positions, however here ORDER MATTERS because there is a ranking.
Let’s say you have A, B, C, D, E, F to run for the board.
If you first chose ABC, that is not the same as CBA because A is the president in the first round and C is the president in the second round. ORDER MATTERS
The following ways could be:
ABC, ACB, BAC, BCA….
THIS IS A PERMUTATION PROBLEM
Using the formula:
There are 120 ways of selecting the board.