Probability:
Week 1: chapter 12+13
Do all assignments, so you can get a bonus of 0.5 on your exam.
Subject in field of probs and stats:
- Descriptive statistics:
- Probability theory:
- (Mathematical) statistics:
Sample space (S) = set of all possible outcomes e.g. S = {0, 1, 2,…., 20}
Roll a dice twice and record both results (as a pair):
Results: (1,1), (1,2), (1,3), …… , (6,6)
Formal notation: S = { (i, j) | i = 1, 2, …., 6 and j = 1, …., 6}
Probabilities: all 36 outcomes of this experiment are equally likely, so P ( (i, j)) = 1/36 for
every outcome (i, j)
Imagine the event ‘’total number in two rolls is 4 can be denoted as:
A = { (1, 3), (2, 2), (3, 1)}
The event A is a subset of sample space S and consists of 3 outcomes
The probability that A occurs is P (A) = 3/36
Venn-diagram: graph events in here
Rules of probability
The complement of A: A^c = {all outcomes not in A}
say: not-A (occurs)
The complement rule
P(A^c) = 1 – P (A)
The intersection A and B
say : A and B (occur)
The union of A and B: A or B
say: A or B or both (occur)
The (general) addition rule= P(A or B) = P(A) + P(B) – P(A and B)
SOMETHING IS INDEPENDENT IF: P(A and B) = P(A) X P(B)
OR P(A|B) = P(A)
, Conditional probability
If A and C are mutually exclusive their intersection is empty, there is nothing in A and C
simultaneously
A and C are mutually exclusive if: P(A and C) = 0
Addition rule for mutually exclusive events:
P( A or C) = P(A) = P(C)
If all n outcomes of an experiment are equally likely and the event A contains k outcomes,
then we get P(A) = k/n
Conditional probability of B for the males(A): P(B|A) =
P(B|A) = probability of B when A is given
So clearly the probability of high grades depends on gender.
Conditional probability and tree diagrams:
Week 1: chapter 12+13
Do all assignments, so you can get a bonus of 0.5 on your exam.
Subject in field of probs and stats:
- Descriptive statistics:
- Probability theory:
- (Mathematical) statistics:
Sample space (S) = set of all possible outcomes e.g. S = {0, 1, 2,…., 20}
Roll a dice twice and record both results (as a pair):
Results: (1,1), (1,2), (1,3), …… , (6,6)
Formal notation: S = { (i, j) | i = 1, 2, …., 6 and j = 1, …., 6}
Probabilities: all 36 outcomes of this experiment are equally likely, so P ( (i, j)) = 1/36 for
every outcome (i, j)
Imagine the event ‘’total number in two rolls is 4 can be denoted as:
A = { (1, 3), (2, 2), (3, 1)}
The event A is a subset of sample space S and consists of 3 outcomes
The probability that A occurs is P (A) = 3/36
Venn-diagram: graph events in here
Rules of probability
The complement of A: A^c = {all outcomes not in A}
say: not-A (occurs)
The complement rule
P(A^c) = 1 – P (A)
The intersection A and B
say : A and B (occur)
The union of A and B: A or B
say: A or B or both (occur)
The (general) addition rule= P(A or B) = P(A) + P(B) – P(A and B)
SOMETHING IS INDEPENDENT IF: P(A and B) = P(A) X P(B)
OR P(A|B) = P(A)
, Conditional probability
If A and C are mutually exclusive their intersection is empty, there is nothing in A and C
simultaneously
A and C are mutually exclusive if: P(A and C) = 0
Addition rule for mutually exclusive events:
P( A or C) = P(A) = P(C)
If all n outcomes of an experiment are equally likely and the event A contains k outcomes,
then we get P(A) = k/n
Conditional probability of B for the males(A): P(B|A) =
P(B|A) = probability of B when A is given
So clearly the probability of high grades depends on gender.
Conditional probability and tree diagrams:
