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## PowerPoint Slideshow about ' Chapter 7 Probability' - tillie

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7.1 From Data to Probability

- In a call center, what is the probability that an
- agent answers an easy call?
- An easy call can be handled by a first-tier agent; a hard call needs further assistance
- Two possible outcomes: easy and hard calls
- Are they equally likely?

7.1 From Data to Probability

- Probability = Long Run Relative Frequency
- Keep track of calls (1 = easy call; 0 = hard call)
- Graph the accumulated relative frequency of easy calls
- In the long run, the accumulated relative frequency converges to a constant (probability)

7.1 From Data to Probability

- The Law of Large Numbers (LLN)
- The relative frequency of an outcome
- converges to a number, the probability of the
- outcome, as the number of observed
- outcomes increases.
- Notes: The pattern must converge for LLN to apply. LLN only applies in the long run.

7.1 From Data to Probability

- The Accumulated Relative Frequency of
- Easy Calls Converges to 70%

7.2 Rules for Probability

- Sample Space
- Set of all possible outcomes
- Denoted by S; S = {easy, hard}
- Subsets of samples spaces are events; denoted as A, B, etc.

7.2 Rules for Probability

- Venn Diagrams
- The probability of an event A is denoted as P(A).
- Venn diagrams are graphs for depicting the relationships among events

7.2 Rules for Probability

- Rule 3: Addition Rule for Disjoint Events
- Disjoint events are mutually exclusive;
- i.e., they have no outcomes in common.
- The union of two events is the collection of outcomes in A, in B, or in both (A or B)

7.2 Rules for Probability

- Rule 3: Addition Rule for Disjoint Events
- Extends to more than two events
- P (E1 or E2or … or Ek) =
- P(E1) + P(E2) + … + P(Ek)

7.2 Rules for Probability

- Rule 4: Complement Rule
- The complement of event A consists of the outcomes in S but not in A
- Denoted as Ac

7.2 Rules for Probability

- Rule 5: Addition Rule
- The intersection of A and B contains the outcomes in both A and B
- Denoted as A ∩ B read “A and B”

7.2 Rules for Probability

- An Example – Movie Schedule

7.2 Rules for Probability

- What’s the probability that the next customer
- buys a ticket for a movie that starts at 9 PM
- or is a drama?

7.2 Rules for Probability

- What’s the probability that the next customer
- buys a ticket for a movie that starts at 9 PM
- or is a drama?
- Use the General Addition Rule:
- P(A or B) = P(9 PM or Drama)
- = 3/6 + 3/6 – 2/6
- = 2/3

7.3 Independent Events

- Definitions
- Two events are independent if the occurrence of one does not affect the chances for the occurrence of the other
- Events that are not independent are called dependent

7.3 Independent Events

- Multiplication Rule
- Two events A and B are independent if the
- probability that both A and B occur is the
- product of the probabilities of the two events.
- P (A and B) = P(A) XP(B)

4M Example 7.1: MANAGING A PROCESS

- Motivation
- What is the probability that a breakdown on an assembly line will occur in the next five days, interfering with the completion of an order?

4M Example 7.1: MANAGING A PROCESS

- Method
- Past data indicates a 95% chance that the
- assembly line runs a full day without breaking
- down.

4M Example 7.1: MANAGING A PROCESS

- Mechanics
- Assuming days are independent, use the
- multiplication rule to find
- P (OK for 5 days) = 0.955 = 0.774

4M Example 7.1: MANAGING A PROCESS

- Mechanics
- Use the complement rule to find
- P (breakdown during 5 days)
- =1 - P(OK for 5 days)
- = 1- 0.774 = 0.226

4M Example 7.1: MANAGING A PROCESS

- Message
- The probability that a breakdown interrupts
- production in the next five days is 0.226. It is wise
- to warn the customer that delivery may be delayed.

7.3 Independent Events

- Boole’s Inequality
- Also known as Bonferroni’s inequality
- The probability of a union is less than or equal to the sum of the probabilities of the events

7.3 Independent Events

- Boole’s Inequality
- Applied to 4M Example 7.1
- P (breakdown during 5 days)
- =P(A1orA2orA3orA4orA5)
- ≤ 0.05 + 0.05 + 0.05 + 0.05 + 0.05
- ≤ 0.25
- Exact answer if the events are independent is 0.226

Best Practices

- Make sure that your sample space includes all of the possibilities.
- Include all of the pieces when describing an event.
- Check that the probabilities assigned to all of the possible outcomes add up to 1.

Best Practices (Continued)

- Only add probabilities of disjoint events.
- Be clear about independence.
- Only multiply probabilities of independent events.

Pitfalls

- Do not assume that events are disjoint.
- Avoid assigning the same probability to every outcome.
- Do not confuse independent events with disjoint events.

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