T3 probability

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QMProbability

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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.

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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

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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.

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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.

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What is Sample Spaces Collection of all possible outcomes.

All six faces of a die:

All 52 cards in a deck:

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Events Simple event

Outcome from a sample space with one characteristic.

e.g.: A red card from a deck of 52 cards

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Visualizing EventsRed cards 26Black cards 26Total cards 52

P(A red card is drawn from a deck) = 26/52= 0.5

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Events Compound event

Involves at least two outcomes simultaneously.

e.g.: An ace that is also red from a deck of cards.

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Visualizing EventsAce Others Total

Red 2 24 26

Black 2 24 26

Total 4 48 52

P(An Ace and Red) = 2/52

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Impossible Events Impossible event

e.g.: One card drawn is ‘Q’ of Club & diamond Also known as ‘Null Event’

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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) = ?

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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=

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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=

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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

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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.

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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’

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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’

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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

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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

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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)

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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)

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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)

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Bayes’s Theorem

P(A l B) = P(B l A) x P(A)

P(B)

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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)

=

=

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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) )

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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

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Example

CD - College degree

NCD – No College degree

RL – Repaid Loan

NRL – Not Repaid Loan

P(RL | CD) = ?

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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

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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)?

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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?

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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?

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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?

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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?

T3 probability

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