CHAPTER 15 PROBABILITY RULES!. THE GENERAL ADDITION RULE Does NOT require disjoint events! P(A U B)...

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CHAPTER 15 PROBABILITY RULES!

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EXAMPLE: USING THE GENERAL ADDITION RULE A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program, and 42% do both. What is the probability that a randomly selected student either lives OR eats on campus?

Transcript of CHAPTER 15 PROBABILITY RULES!. THE GENERAL ADDITION RULE Does NOT require disjoint events! P(A U B)...

Page 1: CHAPTER 15 PROBABILITY RULES!. THE GENERAL ADDITION RULE Does NOT require disjoint events! P(A U B) = P(A) + P(B) – P(A ∩ B) Add the probabilities of.

C H A P T E R 1 5

PROBABILITY RULES!

Page 2: CHAPTER 15 PROBABILITY RULES!. THE GENERAL ADDITION RULE Does NOT require disjoint events! P(A U B) = P(A) + P(B) – P(A ∩ B) Add the probabilities of.

THE GENERAL ADDITION RULE

• Does NOT require disjoint events!

• P(A U B) = P(A) + P(B) – P(A ∩ B)

• Add the probabilities of A and B then subtract the probability of A and B.

Page 3: CHAPTER 15 PROBABILITY RULES!. THE GENERAL ADDITION RULE Does NOT require disjoint events! P(A U B) = P(A) + P(B) – P(A ∩ B) Add the probabilities of.

EXAMPLE:USING THE GENERAL ADDITION RULE

A survey of college students found that 56% live in a campus residence hall, 62% participate in a campus meal program, and 42% do both. What is the probability that a randomly selected student either lives OR eats on campus?

Page 4: CHAPTER 15 PROBABILITY RULES!. THE GENERAL ADDITION RULE Does NOT require disjoint events! P(A U B) = P(A) + P(B) – P(A ∩ B) Add the probabilities of.

EXAMPLE:USING VENN DIAGRAMS

Back to the college students: 56% live on campus, 62% have a meal plan, and 42% do both. Based on a Venn diagram, what is the probability that a randomly selected studenta) Lives off campus and doesn’t have a meal plan?b) Lives on campus but doesn’t have a meal plan?

Page 5: CHAPTER 15 PROBABILITY RULES!. THE GENERAL ADDITION RULE Does NOT require disjoint events! P(A U B) = P(A) + P(B) – P(A ∩ B) Add the probabilities of.

STEP-BY-STEP EXAMPLE

Police report that 78% of drivers stopped on suspicion of drunk driving are given a breath test, 36% a blood test, and 22% both tests.

What is probability that a randomly selected DWI suspect is given…1. A test?2. A blood test or a breath test but not both?3. Neither test?

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CONDITIONAL PROBABILITY: IT DEPENDS…

Conditional Probability: a probability that takes into account a given condition

P(B│A) = the probability of B given that A has occurred

**A usually tells you what to put in the denominator

P(B│A) =

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FINDING CONDITIONAL PROBABILITYEXAMPLE

Our survey found that 56% of college students live on campus, 62% have a campus meal program, and 42% do both. While dining in a campus facility open only to students with meal plans, you meet someone interesting. What is the probability that your new acquaintance lives on campus?

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INDEPENDENCE

Events A and B are independent whenever P(B│A) = P(B).

Are living on campus and having a meal plan independent? Are they disjoint? (Back it up with math)

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INDEPENDENT ≠ DISJOINT

• Disjoint events can NOT be independent!

• Two events could be either independent or disjoint, but not both.

• And they can be NEITHER disjoint nor independent.

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TABLES VS VENN DIAGRAMS

• Step-by-Step Example

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THE GENERAL MULTIPLICATION RULE

• Does not require independence!

• P(A∩B) = P(A) x P(B│A)

• Means the probability of A and B equals the probability of A times the probability of B given A has occurred.

Page 15: CHAPTER 15 PROBABILITY RULES!. THE GENERAL ADDITION RULE Does NOT require disjoint events! P(A U B) = P(A) + P(B) – P(A ∩ B) Add the probabilities of.

EXAMPLE:USING THE GENERAL MULTIPLICATION RULE

A factory produces two types of batteries, regular and rechargeable. Quality inspection tests show that 2% of the regular batteries come off the manufacturing line with a defect while only 1% of the rechargeable batteries have a defect. Rechargeable batteries make up 25% of the company’s production. What is the probability that if we choose 1 battery at random we get…a) A defective rechargeable battery?b) A regular battery and it is not defective?

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DRAWING WITHOUT REPLACEMENT

You just bought a small bag of Skittles. Not that you could know this, but inside are 20 candies: 7 green, 5 orange, 4 red, 3 yellow, and only 1 purple. You tear open one corner of the package and begin eating them by shaking one out at a time. What is the probability that …A) Your first two Skittles are both orange?

B) That none of your first 3 candies is green?

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TREE DIAGRAMS

• A display of conditional events or probabilities that is helpful in organizing our thinking.

• Now lets make a tree diagram with the battery example…