May 02, 2016
Section: 003
Instructor: Nagham Mohammed
Name (Goes by): Nagham
Office: M3 3112
Email: n37mohammad@uwaterloo.ca
Office Hours: 10:30-12:00pm Mondays and Wednesdays
Textbooks: STAT 220/230/240 Course Notes (Spring 2016 Edition)
Can be Acquired: at SCH as courseware
2 Midterms worth 15% each:
Tues, May 31 4:30 - 6:30
Tues, June 28, 4:30 - 6:00
Final Exam
iClicker
Tutorials start next week
May 04, 2016
There are three ways to think about probability:
The subjective probability/ personal-probability definition (probability as a measure of belief).
Number of ways the event can occur/The total number of outcomes
Note: All outcomes are “equally likely”.
Probability of a specific outcome is defined as the proportion of times it occurs over the long run.
Experiment is any action, phenomenon or process that can be infinitely repeated, at least in theory.
Trial is a single repetition of the experiment.
Sample Space of an experiment denoted by S, is the set of all possible distinct outcomes of that experiment.
The Sample Space of rolling 1 die:
S = {1, 2, 3, 4, 5, 6}
2 dice:
S = {1, 2, 3, 4, 5 … 36}
You can get 36 since \(6^2\). There are two six sided dice.
The sample space may be either discrete or non discrete (continuous). A sample space is discrete if it consists of a finite or countably infinite set of simple events.
S= {a1, a2, a3,…} S= {1, 2, 3, 4,…} All positive integers. S= { 1/2, 1/3, 1/4, 1/5…} All rational numbers. We say a set is discrete if the elements in it are ‘separated’.
A sample space is continuous if it contains an interval (either finite or infinite ) of real numbers.
May 06, 2016
What is the sample space if the experiment consist s of measuring (in hours) the life time of a transistor?
S=
The sample space consist of all nonnegative real numbers. (This is a continuous sample space)
An event is any collection (subset) of outcomes contained in the sample space S.
The odds in favour of an event A is defined by \(\frac{P(A)}{P(A^c)} = \frac{P(A)}{1- P(A)}\)
Example
If P(A) = 2 / 3 what are the odds of A (odds in favour of A)?
1 - P(A) =1/3
Odds (A) = \(\frac{P(A)}{1-P(A)}\) = 2 : 1
Let S = { \(a_1, a_2,...a_n\)} be a sample space.
Let P (\(a_i\)), i= 1, 2,…, n be the probabilities to the \(a_i\)’s.
The probability of an arbitrary compound event A can be determined by summing the probabilities of simple events in A.
A = { \(a_1, a2,...a_k\)}
If each simple event has probability \(\frac{1}{n}\) ( i.e. “equally likely”).
P(A) = \(\frac{k}{n}\)
Example
Suppose a 6-sided fair die is rolled, what is the probability of getting an even number ?
Let A = “even number” A = {2, 4, 6}
P (A) = P (2) + P (4) + P (6) = 1/2.
\(\sum\) (Of all i) P (\(a_i\)) =1 , P (S) = 1
For any event A, 0 ≤ P(A) ≤ 1
Probabilities are always between 0 and 1
0: event never happens,
1: event always happens.
If A and B are two events with A ⊆ B
(that is, all of the points in A are also in B)
then P(A) ≤ P(B)
Two events are mutually exclusive
Example
Tossing a coin once, which can result in either heads or tails, but not both.
May 09, 2016
Suppose that two experiments (jobs) are to be performed.
Then if experiment 1 can result in any one of p possible outcomes and there are q possible outcomes of experiment 2.
Then we can do either experiment 1 OR experiment 2 (but not both), in p + q ways.
Example
If there are 5 odd numbers and 9 even and you must pick one number. How many ways you can do that?
5 + 9 = 14 ways.
Suppose that two experiments (jobs) are to be performed.
Then if experiment 1 can result in any one of p possible outcomes and if, for each outcome of experiment 1, there are q possible outcomes of experiment 2, then together (AND) there are p * q possible outcomes of two experiments.
Example A small community consists of 10 women, each whom has 3 children. If one woman and one of her children are to be chosen as mother and child of the year, how many different choices are possible?
3 * 10 = 30 ways
1, 2, 3, 4, 5 (a) How many ways can we choose 2 numbers at random (equally likely) from the above 5, with replacement ?
“With replacement” means that after the first number is picked it is “replaced” in the set of numbers, so it could be picked again as the second number.
(a) 5 * 5 = 25 ways
(b) Find the probability that one number is even?
There are
(2 x 3) + (3 x 2) = 12 ways for this event to occur
P (one number is even) = 12 /25
Example How many different 7-place license plates are possible
(a) If the first 3 places are to be occupied by letters and the final 4
by numbers? (with replacement)
26 x 26 x 26 x 10 x 10 x 10 x 10 = 175,760,000
(b) If repetition among letters or numbers were prohibited?
(without replacement)
26 x 25 x 24 x 10 x 9 x 8 x 7 = 78624000
A permutation, also called an “arrangement number” or “order,” are arrangement of the elements of an ordered list.
Suppose that we have n distinct objects, then the number of ways to arrange these n objects in a row (using each symbol once and only once) is
n! = n (n-1) (n-2)…1
Special definition: 0!=1
Suppose the letters (a, b, c, d, e, f) are arranged at random to form a six-letter word (an arrangement) – we must use each letter once only (.?….)
(a) What is the sample space?
S = { abcdef, abcdfe,…, fedcba }
6!=720
Or using the basic principle of counting 6 x 5 x 4 x 3 x 2 x 1 = 720
Question Ms. Jones has 10 books that she is going to put on her bookshelf. Of these, 4 are mathematics books, 3 are chemistry books, 2 are history books, and 1 is a language book. Ms. Jones wants to arrange her books so that all the books dealing with the same subject are together on the shelf. How many different arrangements are possible?
4! 4! 3! 2! 1! = 6912
May 11, 2016
1, 2, 3, 4, 5
How many ways can we choose 2 numbers from the above 5, without replacement, when the order in which we choose the numbers is important?
\(5^{(2)}\) = \(\frac{5!}{(5-2)!}\) = 20
A pin number of length 4 is formed by randomly selecting and arranging 4 digits from the set {0,1, 2, 3,. . . 9} with replacement.
Find the probability of the event:
A: The pin number is even.
B: The pin number has only even digits.
C: All of the digits are unique.
D: The pin number contains at least one 1.
P(A) = \(\frac{5 * 10^3}{10^4}\)
P(B) = \(\frac{5^4}{10^4}\)
P(C) = \(\frac{10*9*8*7}{10^4}\) = \(\frac{10^{(4)}}{10^4}\)
P(D) = \(\frac{10^4-9^4}{10^4}\)
Any Unordered (order does not matter) sequence of k objects taken from a set of n distinct objects is called a combinations of size k of the objects denoted
\(C^k_n = {n \choose k} = \frac{n!}{k*n-k!}\)
Read (n choose k)
For n and k both non-negative integers with n ≥ k.
Example
Suppose there are 8 students in a group and that 5 of them must be selected to form a basketball team.
(a) How many different teams could be formed?
Use the combination rule with n = 8 and k = 5 as shown below:
\({8 \choose 5}\)
\(\frac{8!}{5!3!}\)
= 56
Example
A committee of 3 is to be formed from a group of 20 people. How many different committees are possible?
\({20 \choose 3}\)
\(\frac{20!}{3!17!}\)
= 1140
May 13, 2016
To determine the number of arrangements of a set of n objects when certain of the objects are indistinguishable (are alike) from each other
There are
Known as Multinomial Coefficient
A chess tournament has 10 competitors, of which 4 are Russian, 3 are from the US, 2 are from Canada, and one from Brazil. If the tournament results lists just the nationalities of the players in the order in which they placed, how many outcomes are possible?
\({10 \choose 4} {6 \choose 3} {3 \choose 2}{1 \choose 1}\) = \(\frac{10!}{4! 3! 2! 1!}\)
May 25, 2016
Two events are independent if they do not influence each other.
Events A and B are independent
if and only if P (A ∩ B) = P (A) P (B)
For any two events A and B with P(B) ˃ 0,
the conditional probability of A given B has occurred is defined by
P (A | B) = \(\frac{P (A ∩ B)}{P (A)}\)
￼P(A | B) + P(\(A^c\) | B) =1 |
P (A ∩ B) = P (A | B)P (A)
Then A and B are independent
if and only if either of the statements is true
P (A) = P (A | B)
Sensitivity is complementary to the false negative rate.
P(T | D) + (\(T^c\) | D) = 1
• Specificity is complementary to the false positive rate.
ccc P(\(T^c\) | \(D^c\))+(T | \(D^c\))= 1
June 08, 2016
random variable
c.d.f. add it up