Download - T3 probability
QM/T3Page 1
QMProbability
QM/T3Page 2
Concept of Probability
You are rolling a die. What are the possible outcomes (value of
the top face) if the die is rolled only once? Outcomes are either 1 or 2 or 3 or 4 or 5
or 6. Chance of occurring ‘1’ is one out of six.
QM/T3Page 3
Concept of Probability Probability that outcome of roll of a die is ‘1’ is
1/6. P(Die roll result = 1) = 1/6
Outcome Probability1 1/62 1/6
3 1/64 1/65 1/66 1/6
Simple EVENT
QM/T3Page 4
Concept of Probability What is the probability that outcome of
roll of a die is an even number. This outcome can occur if the roll result is
2 or 4 or 6. i.e. 3 ways. Number of all possible result is 6. P(Even number) = 3/6 = 0.5 Similarly probability of odd number is 3/6.
QM/T3Page 5
Concept of Probability
P(A) = 3/6 = 0.5 P(B) = 3/6 = 0.5 P(A) + P(B) = 1 P(A) = 1 – P(B)
1 3 5 2 4 6A B
NOTE• A & B are mutually
exclusive.• A & B are mutually
exhaustive.• Sum probability of all outcome
is 1.• Probability is always ≥ 0.
QM/T3Page 6
What is Sample Spaces Collection of all possible outcomes.
All six faces of a die:
All 52 cards in a deck:
QM/T3Page 7
Events Simple event
Outcome from a sample space with one characteristic.
e.g.: A red card from a deck of 52 cards
QM/T3Page 8
Visualizing EventsRed cards 26Black cards 26Total cards 52
P(A red card is drawn from a deck) = 26/52= 0.5
QM/T3Page 9
Events Compound event
Involves at least two outcomes simultaneously.
e.g.: An ace that is also red from a deck of cards.
QM/T3Page 10
Visualizing EventsAce Others Total
Red 2 24 26
Black 2 24 26
Total 4 48 52
P(An Ace and Red) = 2/52
QM/T3Page 11
Impossible Events Impossible event
e.g.: One card drawn is ‘Q’ of Club & diamond Also known as ‘Null Event’
QM/T3Page 12
Joint Probability P(An ‘Ace’ and ‘Red’ from a deck of cards)
Ace Others TotalRed 2 24 26
Black 2 24 26
Total 4 48 52
A = ‘Ace’B = ‘Red’
P(A & B) = ?
QM/T3Page 13
Joint Probability The probability of a joint event, A and B: P(A and B) = P(A ∩ B)
A BA & B
No. of outcomes from A and B
Total No. of possible outcomes=
QM/T3Page 14
Compound Probability Probability of a compound event, A or B: P(A or B) = P(A U B)
A BA & B
No. of outcomes from A only or B only or Both
Total No. of possible outcomes=
QM/T3Page 15
Compound Probability P(A or B) = P(A U B)
= P(A) + P(B) – P(A ∩ B)
All Ace All Red4 262
452
2652
252
= + - =713 Addition Rule
QM/T3
Compliment Set In roll of a die
Set of all outcomes = {1,2,3,4,5,6}. If A = {1} then Ac has {2,3,4,5,6}. If B = Set of even outcomes = {2, 4, 6} then Bc
has {1,3, 5}. In a class if A = Set of students that have
passed, then Ac is = Set of students that have not passed.
Page 16
QM/T3Page 17
Computing Joint Probability
Event
EventTotal
B B c
A 2 2 4Ac 24 24 48Total 26 26 52
A = Card drawn from deck is ‘Ace’Ac = Card drawn from deck is not ‘Ace’B = Card drawn from deck is ‘Red’B c = Card drawn from deck is not ‘Red’
QM/T3Page 18
Computing Joint Probability
EventEvent
TotalB B c
A A ∩ B A ∩ B c AAc Ac ∩ B Ac ∩ B c Ac
Total B B c
A = Card drawn from deck is ‘Ace’Ac = Card drawn from deck is not ‘Ace’B = Card drawn from deck is ‘Red’B c = Card drawn from deck is not ‘Red’
QM/T3Page 19
Computing Joint Probability
Event
EventTotal
B B c
A P(A ∩ B) P(A ∩ B c) P(A)
Ac P(Ac ∩ B) P(Ac ∩ B c) P (Ac)
Total P(B) P(B c) 1
QM/T3Page 20
Conditional Probability Finding probability of an event A, given that
event B has occurred. This means that we need to find out probability
of occurrence of ‘Ace’ given that the card drawn is ‘Red’.
Event
EventTotal
B B c
A 2 2 4Ac 24 24 48
Total 26 26 52
=226
QM/T3Page 21
Conditional Probability The probability of event A given that event
B has occurred
= P(A ∩ B)
P(B)This is known as
‘Conditional Probability’ and denoted as:
P(A l B)
QM/T3Page 22
Multiplication Rule
P(A ∩ B) = P(A ∩ B)P(B)
x P(B)
= P(A l B) x P(B)
= P(B l A) x P(A)
QM/T3Page 23
Statistical Independence Events A and B are independent if the
probability of one event, A, is not affected by another event, B
P(A l B) = P(A) P(B l A) = P(B) P(A and B) = P(A) x P(B)
QM/T3Page 24
Bayes’s Theorem
P(A l B) = P(B l A) x P(A)
P(B)
QM/T3Page 25
Bayes’s Theorem
P(A l B) = P(B l A) x P(A)
P(B)
P(B l A) x P(A)
P(B ∩ A) + P(B ∩ Ac)
P(B l A) x P(A)P(B l A) x P(A) + P(B l Ac) x P(Ac)
=
=
QM/T3Page 26
Bayes’s Theorem (General)
P(Bi l A) = P(A l Bi) x P(Bi)
P(A l B1).P(B1) +…+ P(A l Bk).P(Bk) )
QM/T3Page 27
Fifty percent of borrowers repaid their loans. Out of those who repaid, 40% had a college degree. Ten percent of those who defaulted had a college degree. What is the probability that a randomly selected borrower who has a college degree will repay the loan?
Example
Repaid loan Not repaid loan Total
College degree
No college degree
Total 0.5
0.2
0.5
0.05
0.25
0.3 0.45
0.751.0
QM/T3Page 28
Fifty percent of borrowers repaid their loans. Out of those who repaid, 40% had a college degree. Ten percent of those who defaulted had a college degree. What is the probability that a randomly selected borrower who has a college degree will repay the loan?
Example
CD - College degree
NCD – No College degree
RL – Repaid Loan
NRL – Not Repaid Loan
P(RL | CD) = ?
QM/T3Page 29
Fifty percent of borrowers repaid their loans. Out of those who repaid, 40% had a college degree. Ten percent of those who defaulted had a college degree. What is the probability that a randomly selected borrower who has a college degree will repay the loan?
Example
CD - College degreeNCD – No College degreeRL – Repaid LoanNRL – Not Repaid Loan
P(RL | CD) = ?
P(CD | RL) P(RL)
P(CD | RL) P(RL) + P(CD | NRL) P(NRL)
0.4 0.5
0.4 0.5 0.1 0.5
QM/T3Page 30
Class Exercise1. P(A) = 0.25, P(B) = 0.4, P(A|B) = 0.15
Find out P(AUB).2. Probability of two independent events A
and B are 0.3 and 0.6 respectively. What is P(A∩B)?
3. P(A∩B) = 0.2 ; P(A∩C) = 0.3 ;P(B|A) + P(C|A) = 1; What is P(A)?
QM/T3Page 31
Class Exercise4. A jar contains 6 red, 5 green, 8 blue and
3 yellow marbles.a) What is the probability of choosing a red
marble?5. You are tossing a coin three times.
a) What is the probability of getting two tails?b) What is the probability of getting at least 2
heads?
QM/T3Page 32
Class Exercise6. A plant has 3 assembly lines that
produces memory chips. Line1 produces 50% of chips (defective 4%), Line2 produces 30% of chips (defective 5%), Line3 produces the rest (defective 1%). A chip is chosen at random from produced lot.
a) What is the probability that it is defective?b) Given that the chip is defective, what is the
probability that it is from Line2?
QM/T3Page 33
Class Exercise7. An urn initially contains 6 red and 4
green balls. A ball is chosen at random and then replaced along with two additional ball of same colour. This process is repeated.
a) What is the probability that the 1st and 2nd ball drawn are red and 3rd is green?
b) What is the probability of 2nd ball drawn is red?
QM/T3Page 34
Class Exercise8. Two squares are chosen at random on a
chessboard. What is the probability that they have a side in common?
9. An anti aircraft gun can fire four shots at a time. If the probabilities of the first, second, third and the last shot hitting the enemy aircraft are 0.7, 0.6, 0.5 and 0.4, what is the probability that four shots aimed at an enemy aircraft will bring the aircraft down?