ch. 11 conditional probability p(a|b) = p(a and b)/ p(b), assuming p(b) > 0 p(b|a) = p(a and b)/...

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Ch. 11 Conditional Probability • P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 • P(B|A) = P(A and B)/ P(A), P(A) > 0

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Page 1: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Ch. 11 Conditional Probability

• P(A|B) = P(A and B)/ P(B),

assuming P(B) > 0

• P(B|A) = P(A and B)/ P(A), P(A) > 0

Page 2: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Examples

• In a monthly report, the local animal shelter states that it currently has 24 dogs and 18 cats available fop adoption. Eight of the dogs and six of the cats are male. Find each of the following probabilities.

• a) The pet is male given that it is a cat• b) The pet is a cat, given that it is female• c) The pet is dog given that it is a female

Page 3: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Example

• You draw a card at random from a standard deck of 52 cards. Find the following probabilities.

• a) The card is a heart, given that it is red• b) The card is red, given that it is a heart• c) The card is an ace, given that it is red• d) The card is a queen, given that it is face card

Page 4: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Ch. 11 Conditional Probability

Ch. 14/15 From randomness to Probability

Page 5: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Example• 3. The Masterfoods Company says that before the

introduction of purple, yellow candies made up of 20% of their plain M&M’s, red another 20% and orange, blue and green each made up 10%. The rest were brown.

• a) If you pick an M&M at random, what’s the probability that• i) it is brown• ii) it is yellow or orange?• iii) it is not green?

Page 6: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Examples

• 4. If you pick three M&M’s in a row, what s the probability that

• a) they are all brown? • b) the third one is the first one that is red?• c) none are yellow?• d) at least one is green

Page 7: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Example

• The probabilities that an adult American man has high blood pressure and/or high cholesterol are shown in the table

HIGHBlood Pressure

OK

High Cholesterol

0.11 0.21

OK 0.16 0.52

Page 8: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Example• a) What is the probability that a man has both

conditions? • b) What is the probability that a man has high blood

pressure?• c) What is the probability that a man with high blood

pressure has high cholesterol? • d) What is the probability that a man has high blood

pressure if it is known he has high cholesterol?• e) Are high blood pressure and high cholesterol

independent?

Page 9: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Example

• A poll conducted by the University of Montana classified respondents by gender and political party as shown in the table. Is party affiliation independent of gender?

D R I

Male 36 45 24

Female 48 33 16

Page 10: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Example

D R I

M .18 .22 .12 .52

F .24 .16 .08 .48

.42 .38 .20 1

Page 11: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Chapter 11 Conditional Probability and Tree Diagram

1% of employees in a company are drug

users. A test was developed to identify the

drug users. If a person is a drug user, then

the test returns positive result 98% of the

time. If a person is not a drug user, the test

returns negative result 97% of the time.

Will you recommend this test to be

administered at your company?

Page 12: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Conditional Probability and Tree Diagrams

Leah is flying from Boston to Denver with a connection in

Chicago. The probability her first flight leaves on time is

0.15. If the flight is on time, the probability her luggage

will make the connection flight in Chicago is 0.95, but if the

first flight is delayed, the probability that the luggage will

make it is only 0.65.

a) What is the probability that her luggage arrives on time?

Answer: 0.695b) Are the first flight leaving on time and the luggage making the

connection independent events? NO

Page 13: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Conditional Probability and Tree Diagrams

A package will be picked up by one of 3 deliveryvans, depending on which is nearest. The choicesare equal for any of the vans to make the pick up.Past record for each van’s success in deliveringpackages are given as follows: • Prob(on time delivery for van 1) = .95 • Prob(on time delivery for van 2) =.96 • Prob(on time delivery for van 3) = .90.

• What is the probability a package will be delivered on time? ANS 0.94

• Is package deliver independent of delivery van? NO• If package is delivered on time, what is the probability that it

was delivered by Van 1? ANS 0.338

Page 14: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Bayes’ Theorem-Reversing the Conditioning

)()|()()|(

)()|()|(

cBPcBAPBPBAP

BPBAPABP

Page 15: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

At a gas station, 40% of customers pump regular gas, 35% pump midgrade gas, and 25% pump premium grade gas. Of those who pump regular, 30% pay at least $30. Of those who pump midgrade, 50% pay at least $30. And of those who pump premium, 60% pay at least $30.

a) What is the probability that the next customer will pay more than $30? ANS 0.445

b) If the next customer pays more than $30, what is the probability that he pumped regular gas? ANS 0.269

Page 16: Ch. 11 Conditional Probability P(A|B) = P(A and B)/ P(B), assuming P(B) > 0 P(B|A) = P(A and B)/ P(A), P(A) > 0

Reversing the Condition

A firm that specializes in filing out incometax forms has two types of accounts• Large(gross income greater than $50000)• Small(gross income less than $50000)They find that 5% of the large accounts and2% of the small accounts are audited by theIRS. If 40% of their accounts are largeaccounts, and an account is audited, what isthe probability that it is large? ANS 0.625