statistics and probability course, Chapter 3 Elementary Probability

1. CHAPTER 3
ELEMENTARY PROBABILITY

2. 2-2
DEFINITION OF SOME PROBABILITY TERMS
 Probability Experiment: is a process that leads to well-defined results which
can’t be predicted in advance called outcomes.
 Outcome/sample point: is the result of a single trial of probability experiment.
 Sample space (S): is a set of all possible outcomes.
 Event: is a subset of sample space.
 simple, compound and null event
Examples:
Prob. Experiment Outcome Sample space(S) Events
Rolling a die once 1 or 2 or … 6 S={1,2,3,4,5,6} • Odd number : A={1,3,5},
• Even number: B= {2,4,6}
• Number >6 appears: C={} or Ø
Tossing a coin twice HT, TH, HH or TT S={HT,TH, HH ,TT} • At least 2 heads: A={HH}
• Head in first toss: B={HT,HT}

3. 2-3
DEFINITIONS OF SOME PROBABILITY TERMS CONT’D…..
Union, Intersection and complement of events
 A u B : outcomes that belong to either A, B or both A and B.
 A n B: outcomes that belong to both A and B.
 Ac : outcomes that do not belong to event A.
 Note that: Sc=Ø
Example : A fair die is rolled once. Let A = {1,3,5} and B = {2,4,6}
• A u B= {1,3,5,2,4,6}
• A n B=Ø
• Ac = {2,4,6}

4. 2-4
DEFINITIONS OF SOME PROBABILITY TERMS CONT’D…..
Equally Likely Outcomes: outcomes which have
the same chance of occurrence.
Mutually Exclusive Events: events with no common
outcome.
 So two events A and B are said to be mutually exclusive if A n B=Ø.
In general, events A1 , A2 ….. Aj …….. Ak, are said to be
mutually exclusive events if Ai n Aj = Ø for all i≠j
Exhaustive events: are mutually exclusive events in which
their union covers the whole sample space.

5. 2-5
Exercise :
1) A fair coin is tossed three times.
a) List all the possible outcomes.
b) List outcomes contained in the event that;
i. no heads appear
ii. two heads appear
iii. at least two heads appear
2) List all possible outcomes contained in two rolls of a fair
die.

6. 2-6
BASIC PRINCIPLES OF COUNTING CONT’D….
i. Multiplication Rule: If an experiment consists of
r steps of which the first can be done in n1 ways,
for each of these the second can be done in n2
ways and rth step can be done in nr ways, then the
whole choice can be done in
n1 ×n2 × n3 × …. × nr ways

7. 2-7
BASIC PRINCIPLES OF COUNTING CONT’D….
Solution:
Let’s use the multiplication rule:
i) If repetition is allowed : 𝟐𝟔 × 𝟐𝟔 × 𝟐𝟔 × 𝟏𝟎 × 𝟏𝟎 × 𝟏𝟎 × 𝟏𝟎 = 175,760,000
possible plates
ii) If repetition is prohibited: 26 × 25 × 24 × 10 × 9 × 8 × 7 = 78, 624, 000
possible plates.
Example : How many different 7-place license plates are
possible if the first 3 places are to be occupied by letters and the
final 4 by numbers?
i)If repetition among letters or numbers is allowed?
ii) If repetition among letters or numbers is prohibited?

8. 2-8
BASIC PRINCIPLES OF COUNTING CONT’D….
ii. Permutation : An arrangement of n objects with regard to order.
-The number of permutations of n objects taking r objects at
a time is:
nPr =
𝒏!
(𝒏 − 𝒓)!
- The number of distinct permutation of n objects in which 𝑛1
are alike, 𝑛2 are alike up to 𝑛𝑘 is give by:
𝒏!
𝒏𝟏!×𝒏𝟏×⋯×𝒏𝒌
iii. Combination: selection of objects without regard to order.
-The number of combinations of n objects taking r at a time without
regard to order :
nCr =
𝒏!
𝒓!(𝒏−𝒓)!

9. 2-9
BASIC PRINCIPLES OF COUNTING CONT’D…..
Example : Find the number of permutations and combinations, if we take two
letters at a time from the letters A, B, C, and D.
Solution: n=4 and r=2
Permutations: Combinations
4P2= 4C2 =
AB BA CA DA
AC BC CB DB
AD BD CD DC
AB BC
AC BD
AD DC
6
!
2
!
2
!
4

 
12
!
2
4
!
4



10. 2-10
BASIC PRINCIPLES OF COUNTING CONT’D…..
Example : From a class of 5 women and 7 men, how many different committees
consisting of 2 women and 3 men can be chosen?
Solution:
5C2 × 7C3 = 350
35
10
!
4
!
3
!
7
!
3
!
2
!
5







11. 2-11
EXERCISE ON BASIC PRINCIPLES OF COUNTING CONT’D…..
Exercise:
1. Among 15 clocks there are two defectives. In how many ways
can an inspector chose three of the clocks so that:
a) None of the defective clock is included. Answer: 286 ways
b) One of the defective clock is included. Answer: 156 ways

12. 2-12
APPROACHES TO MEASURE PROBABILITY
1. Classical Approach:
Assumes equally-likely outcomes.
𝑷 𝑨 =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑡𝑜 𝑒𝑣𝑒𝑛𝑡 𝐴
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
=
𝒏(𝑨)
𝒏(𝑺)

13. 2-13
APPROACHES TO MEASURE PROBABILITY
Examples:
1. A fair coin is tossed once. Let A be the event “head” appears. Then,
P(A)=
𝒏(𝑨)
𝒏(𝑺)
=1/2
2. A box of 80 items consists of 30 defective and 50 non defective items. If 10 of
these items are selected at random, what is the probability all will be defective?
 Let A be the event that all 10 items will be defective.
 The total # of ways event A occurs: n(A)= 30C10 × 50C0
 Total selection: n(s)= 80C10
Then, 𝑃 𝐴 =
𝒏(𝑨)
𝒏(𝑺)
= 30C10 × 50C0
80C10
=0.00001825

14. 2-14
APPROACHES TO MEASURE PROBABILITY CONT’D….
2. Relative frequency /Empirical approach
 Based on the relative frequencies of outcomes from repetition of expt.
NA = # of times event A occurs
N = # of times random experiment is repeated
 Probability of getting a head in N tosses of a coin:
3. Subjective Approach
 Assigns probability on the basis of individuals judgment/view.
P(A)=lim
𝑁→∞
𝑁𝐴
𝑁
number of
tosses(N)
Number of
heads(𝑵𝑨)
P(A)
Count Buffon 4,040 2,048 0.5069
John Kerrich 10,000 5,067 0.5067
Karl Pearson 24,000 12,012 0.5005

15. 2-15
APPROACHES TO MEASURE PROBABILITY CONT’D….
4. Axiomatic Approach
 Axioms of Probability:
i. Let S be a sample space. Then, P(S)=1
ii. For any event A, 0 ≤ P(A) ≤ 1.
iii. If A and B are mutually exclusive, then P(AUB)= P(A) +P(B).
 In general, if A1 , A2 ….. are mutually exclusive events, then
P(A1 U A2 U…..)=P(A1) + P(A2)+…..
Remarks:
P(Ac)= 1-P(A)
P(Ø)=P(Sc)= 0
P(A)= P(AnBc) + P(AnB)

16. 2-16
CONDITIONAL PROBABILITY AND INDEPENDENCE
For any two events A and B;
 The probability of event A given B is given as:
where, P (B)> 0
 Equivalently, probability of event B given A is given as:
P (B/A)= where, P (A) > 0
)
(
)
(
A
P
AnB
P
)
(
)
(
)
(
B
P
B
A
P
B
A
P 

Remark: P (Ac/B)=1- P (A/B) and P (Bc/A)=1- P (B/A)

17. 2-17
CONDITIONAL PROBABILITY AND INDEPENDENCE
INDEPENDENT EVENTS
 Two events A and B are said to be independent, if
P(A/B)=P(A) and P(B/A)=P(B).

18. 2-18
CONDITIONAL PROBABILITY CONTD.
Example : 125 employees of a certain factory are given a
performance test and are divided in to two groups as
those with good performance(G) and those with poor
performance (P) the result is given below. Find P(M/G)
and P(F/P)?

19. 2-19
CONDITIONAL PROBABILITY CONTD.
Solution :
P(M/G) i.e the probability that the selected is person is male given that
it has a good performance is:
𝑷 𝑴 𝑮 = 𝑷(𝑴∩𝑮)
𝑷(𝑮)
= 𝟔𝟎 𝟏𝟐𝟓
𝟖𝟓 𝟏𝟐𝟓
=
𝟏𝟐
𝟏𝟕
=𝟎.𝟕𝟏
P(F/P) i.e the probability of a person is female given that it has a poor
performance is :
𝑷 𝑭 𝑷 = 𝑷(𝑭∩𝑷)
𝑷(𝑷)
= 𝟏𝟓 𝟏𝟐𝟓
𝟒𝟎 𝟏𝟐𝟓
=𝟎.𝟒𝟑

20. 2-20
SOME RULES OF PROBABILITY
i. Addition rule
 Probability that either A or B or both A and B occur:
Example : A fair coin is tossed twice. Let event A={HH,HT} and B={HH,TH}. Find
P(AUB).
Solution:
S = {HH,HT,TH,TT} , P(A)=½ P(B)= ½ P(AnB)= ¼
P(AUB) =P(A) + P(B) - P(AnB)
= ½ + ½ - ¼
= ¾
P(A u B) = P(A) + P(B) – P(A n B)

21. 2-21
SOME RULES OF PROBABILITY
ii. Multiplication rule
 Let A and B be any events. Then, the probability that both A and B occur is
given as:
 If A and B are independent events the multiplication rule is reduces to:
P(A n B)= P(A/B)* P(B)
=P(B/A)*P(A)
P(AnB)=P(A)*P(B)

22. 2-22
SOME RULES OF PROBABILITY
Example :A box contains 25 items, of which 5 are defective. Two items
are randomly picked in sequence. What is the probability that both
the items are defective? If the items are picked
i. Without replacement?
ii. With replacement
Solution:
Let A=event that the first item is defective
B=event that the second item is defective
Then
i. P(AnB)= P(A)*P(B/A)
= 5/25*4/24=0.03
ii. P(AnB)= P(A)*P(B)
=5/25*5/25=0.04

23. 2-23
SOME RULES OF PROBABILITY
Example : A coin is tossed and a die is rolled. Find probability of
getting head and number 4.
Solution:
Let A=the event that head appears in tossing the coin
B=the event that number 4 appears in rolling the die
We want to find P(AnB). Since A and B are independent:
P(AnB) =P(A)*P(B)
=(1/2)*(1/6)
=1/12

24. 2-24
Example: Let A and B be two events associated with an experiment and suppose
that P(A)=0.4 while P(AUB)=0.7. Let P(B)=P
a) For what choice of P are A and B mutually exclusive?
b) For what choice of P are A and B independent?
a) If A and B are mutually exclusive, then P(AnB)=P ∅ = 0
Using addition rule:
P(AuB)= P(A) + P(B) –P(A nB)
== > 0.7=0.4 + P-0
== > P=0.3
b) If A and B are independent , then P(A nB)=P(A) ×P(B)
Using addition rule we have
P(AuB)= P(A) + P(B) –P(A) ×P(B)
0.7=0.4 + P- 0.4 ×P
== > P=0.5
SOME RULES OF PROBABILITY…

25. 2-25
EXERCISE