Prob-Lecture-7-BRIS-FA2024-u-06102024-114300am.pdf

Probability lecture by haad

1. 1GSC 122 Haad Akmal

2. “Probability and Statistics”
Course Code: GSC 122
Credit Hours: 3(3+0)
Prerequisite:
Course Instructor:
Dr. Haad Akmal,
AP,
RIS @ BUIC.
GSC 122 Haad Akmal 2

3. Probability of an Event
3GSC 122 Haad Akmal

4. Lecture Outline:
Probability of an Event.
Different Approaches to Probability.
Axioms of Probability.
Examples.
4GSC 122 Haad Akmal

5. Probability:
“Pakistan will probably win the cricket
match,”
“I have a fifty-fifty chance of getting an
even number when a die is tossed,”
“The university is not likely to win the
football game tonight,”
“Most of our graduating class will likely be
married within 3 years”.
5GSC 122 Haad Akmal

6. Probability of an Event:
The likelihood of the occurrence of an event
resulting from such a statistical experiment is
evaluated by means of a set of real numbers,
called weights or probabilities, ranging from
0 to 1.
To every (sampling) point in the sample space,
a probability is assigned such that the sum of
all probabilities is 1.
Note:
◼For points outside the sample space (i.e., for
events that cannot possibly occur) a probability of
0 is assigned. Example?
6GSC 122 Haad Akmal

7. Probability of an Event: Cont…
The probability of an event A is the sum of the
weights/probabilities of all the sample
points in event A.
0≤??????(??????)≤1, ??????(∅)=0, ????????????????????????(??????)=1.
◼S is the sample space from which A is taken.
If A
1, A
2, A
3, . . . is a sequence of mutually
exclusive events, then
??????(??????
1 ∪ ??????
2 ∪ ??????
3 ∪…)=??????(??????
1)+??????(??????
2)+??????(??????
3)…
7GSC 122 Haad Akmal

8. Probability of an Event:
Example 1
A coin is tossed twice. What is the probability
that at least 1 head occurs?
S = {HH, HT, TH, TT}
If the coin is balanced, each of these
outcomes is equally likely to occur. Then
probability(or weight) denoted as p (or ω)
of each sample point:
p + p + p + p = 4 p = 1, or p = 1/4.
If A represents the event of at least 1 head
occurring, then A = {HH, HT, TH}:
P(A) = ¼ + ¼ + ¼ = ¾
8GSC 122 Haad Akmal

9. Probability of an Event:
Example 2a
A dice is loaded in such a way that an even
number is twice as likely to occur as an odd
number.
The total probability for odd numbers is 3p.
The total probability for even numbers is 3(2p)=6p.
Since the sum of probabilities for all outcomes
must equal 1:
◼3p + 6p = 1
◼9p = 1
◼p = 1/9
◼Thus, the probability of an odd number is p=1/9, and the
probability of an even number is 2p =2 (1/9), p = 2/9.
9GSC 122 Haad Akmal

10. Probability of an Event:
Example 2b
If E is the event that a number less than 4 occurs
on a single toss of the die, find P(E):
{1, 2, 3}:
The probabilities of these outcomes are:
For outcome 1 on the dice roll = P(1) = p = 1/9
For outcome 2 on the dice roll = P(2) = 2p = 2 (1/9)
= 2/9
For outcome 3 on the dice roll = P(3) = p = 1/9
So, the probability of event E is the sum of these
probabilities:
P(E) = P(1) + P(2) + P(3) = 1/9 + 2/9 + 1/9 = 4/9.
10GSC 122 Haad Akmal

11. Probability of an Event:
Example 2c
If A is the event that an even number turns up, and
B is an event that a number is divisible by 3. Find
P(A U B) & P(A ∩ B):
A = {2, 4, 6} and B = {3,6}
From part 2b: Since the sum of probabilities for all
outcomes must equal 1: p = 1/9
A U B = {2,3,4,6}
P(A U B) = 2/9 + 1/9 + 2/9 + 2/9 = 7/9
A ∩ B = {6}
P(A ∩ B) = 2/9
11GSC 122 Haad Akmal

12. Popup Quiz # 1:
Q: In example 2, why was the probability of an
outcome 1/9?
A: Because of the condition given in the question
(loaded dice in favor of an even numbered
outcome). Otherwise, it would have been 1/6.
For a non loaded dice, where the outcomes are
equally likely to occur:
p(odd) + p(even) = 1
3p + 3p = 1
6p = 1
p = 1/6
12GSC 122 Haad Akmal

13. Probability for Equally Likely
Outcomes:
If an experiment can result in any one of N
different equally likely outcomes, and
If exactly n of these outcomes correspond to
event A, then the probability of event A is
????????????=
??????
??????
Note: n are usually the desired outcomes of an
experiment.
13GSC 122 Haad Akmal

14. Probability of an Event:
Example 3
A Probability class for a program consists of 25
pre-medical (M), 10 pre-engineering (E), 8 ics
(C) and 10 (A) arts students. If a student is
randomly selected by the instructor to answer
a question, find the probability that the student
chosen is
(a) a pre-medical background student.
(b) a pre-engineering or an ics background
student.
14GSC 122 Haad Akmal

15. Probability of an Event:
Example 3
(a) a pre-medical background student.
As total number of students = 25+10+8+10 =
53 (N)
Since 25 (n) of the 53 students are from pre-
medical background, the probability (P(M)) of
selecting such a student is:
????????????=
25
53
15GSC 122 Haad Akmal

16. Probability of an Event:
Example 3
(b) a pre-engineering or an ics background
student.
Since 18 (n) of the 53 students are from pre-
engineering or ics backgrounds, So P(E U C) would be:
????????????∪??????=
18
53
16GSC 122 Haad Akmal

17. Probability of an Event:
Example 4
In a poker hand consisting of 5 cards, find the
probability of holding 2 aces and 3 jacks.
Total number of ways (N) to choose 5 cards from a
deck of 52 cards is found using combination formula:

52
5
=
52!
5!(52−5)!
=2,598,960
Number of ways to select 2 aces (out of 4 in deck):

4
2
=
4!
2!(4−2)!
=6
Number of ways to select 3 jacks (out of 4 in deck):

4
3
=
4!
3!(4−3)!
=4
17GSC 122 Haad Akmal

18. Probability of an Event:
Example 4
Using multiplication rule, there are n =
(6)(4) = 24 hands with 2 aces and 3 jacks.
Then probability is the ratio of desired
outcomes (n) to the total number of
possible hands (N):
P= 24/ 2,598,960
18GSC 122 Haad Akmal

19. Popup Quiz # 2:
Q: In example 4, why combination is used instead
of permutation?
A: Because while dealing cards to a hand in the
game (ex 4) of cards, the order of cards does not
matter.
19GSC 122 Haad Akmal

20. Approaches to Probability:
Indifference Approach:
◼Assumes equal likelihood for each outcome.
Subjective Probability:
◼Past performance in games (e.g., cricket).
◼Historical records (e.g., sporting world cups).
◼Personal beliefs and intuition.
◼When prior information and opinions differ,
subjective probability becomes crucial.
20GSC 122 Haad Akmal

21. Approaches to Probability:
When outcomes of an experiment are not equally
likely, probabilities must be assigned based on:
◼Prior knowledge.
◼Experimental evidence.
Dealing with unbalanced outcomes:
◼Estimate probabilities by performing the experiment
multiple times.
◼Use relative frequency to determine true
probabilities.
21GSC 122 Haad Akmal

22. Relative Frequency:
Relative Frequency is based on statistical experiments
rather than subjective opinions. It is essential for
repeatable experiments in science and engineering.
Relative Frequency = f/n
◼f = number of times an outcome was observed in an
experiment (or trial).
◼n = number of times an experiment was performed.
Probability is the relative frequency over an infinite
number of experiments (or trials).
22GSC 122 Haad Akmal
Number of
Coin Tosses
Number of
Heads
Relative
Frequency
5 4 4/5 = 0.8
10 6 6/10 = 0.6
50 23 23/50 = 0.4

23. Formulae for Probability of an
Event:
23GSC 122 Haad Akmal

24. Axioms of Probability:
Probability of an event “A” is non-negative.
◼Pr[A] >= 0
Probability of an outcome belonging to
sample space is 1.
◼Pr[S] = 1
Probability of sum of mutually exclusive
events is equal to sum of probabilities of
events.
◼Pr[A U B] = Pr[A] + Pr[B]
◼For Pr[A∩B] = 0
24GSC 122 Haad Akmal