Chapter 14 Probability Rules!

Chapter 14: Probability Rules

Review of Notation

• $\cap$ represents the intersection of events (both events occur).
• $\cup$ represents the union of events (either event or both occur).

The General Addition Rule

• The addition rule for disjoint events (from the previous chapter) only applies when events A and B are disjoint.
• If events are not disjoint, the earlier addition rule double counts the probability of both A and B occurring.
• The General Addition Rule is:
  P(A \cup B) = P(A) + P(B) - P(A \cap B)
• This rule accounts for the overlap between events.

"or"

• In general English, “or” is often exclusive (one or the other, but not both).
• In statistics and math, “or” is inclusive (A or B or both).

Examples of General Addition Rule

• Example 1: 63% of students have taken PreCalculus, 26% have taken AP Statistics, and 11% have taken both.
  • Percentage of students who have taken AP Statistics but not PreCalculus: 15%.
  • Percentage of students who have taken AP Statistics or PreCalculus: 63% + 26% - 11% = 78%.
  • Percentage of students who have taken neither AP Statistics nor PreCalculus: 22%.
• Example 2: Real estate ads suggest that 64% of homes have garages, 21% have swimming pools, and 17% have both.
  • Probability that a house has a garage or swimming pool: 64% + 21% - 17% = 68%.
  • Probability that a house has a pool but no garage: 21% - 17% = 4%.
  • Probability that a house does not have a pool nor garage: 32%.

Conditional Probability

• Conditional probability considers the probability of an event given a specific condition.
• Notation: P(B|A), read as “the probability of B given A.”

Example of Conditional Probability

• Jane is flying to Hawaii, switching planes in Los Angeles. The probability her first flight leaves on time is 0.15. If the first flight is on time, the probability her luggage makes the connecting flight is 0.8. If the first flight is delayed, the probability her luggage makes it is 0.65.
  • P(\text{luggage} | \text{delayed}) = 0.65
  • P(\text{luggage} | \text{on time}) = 0.80
• To find the probability of event B given event A, restrict attention to outcomes in A and find the fraction of those outcomes where B also occurred.

Conditional Probability Example

• 63% of students have taken PreCalculus, 26% have taken AP Statistics, and 11% have taken both.
  • Probability that a student who has taken AP Statistics has also taken PreCalculus:
    P(\text{PreCalculus} | \text{AP Statistics}) = \frac{P(\text{PreCalculus} \cap \text{AP Statistics})}{P(\text{AP Statistics})} = \frac{0.11}{0.26} = 0.423

The General Multiplication Rule

• When events A and B are independent, the multiplication rule from Chapter 14 can be used: P(A \cap B) = P(A) * P(B).
• When events are not independent, the General Multiplication Rule is needed.
• The General Multiplication Rule, derived from the conditional probability formula, is:
  P(A \cap B) = P(A) * P(B|A)
  or
  P(A \cap B) = P(B) * P(A|B)

General Multiplication Rule Example

• A student has tickets to Spider-Man but has to perform in an orchestra concert. There is a 40% chance the concert goes long. If the concert does not go long, the student has a 90% chance of making the movie. If the concert goes long, the student has a 65% chance of making the movie.
  • Probability that the concert goes long and the student makes the movie: 0.4 * 0.65 = 0.26
  • Probability that the concert does not go long and the student makes the movie: 0.6 * 0.9 = 0.54

Independence

• Independence of two events means the outcome of one does not influence the probability of the other.
• Events A and B are independent if P(B|A) = P(B). Equivalently, P(A|B) = P(A).

Independence Example

• A survey found that 52% of residents have a home phone, 68% have a cell phone, and 39% have both.
  • If a person has a home phone, the probability they also have a cell phone: 39/52 = 0.75.
  • Are having a home phone and a cell phone independent events? No, because P(cell) = 0.68 and P(cell | home) = 0.75.

Independent vs. Disjoint

• Disjoint events cannot occur together.
• Independent events mean knowing the outcome of one doesn’t provide new information about the outcome of the other.
• Disjoint and independent are not the same.

Drawing Without Replacement

• Sampling without replacement means once an individual is drawn, it's not returned to the pool.
• This is a form of conditional probability.

No Replacement Examples

• If I have a standard deck of cards (52), what are the chances that I get dealt three aces in a row?
  • 4/52 * 3/51 * 2/50 = 0.000181
• If I have a standard deck of cards, what are the chances that I get dealt four red cards in a row?
  • 26/52 * 25/51 * 24/50 * 23/49 = 0.0552

Replacement Example

• If I have a fun-size pack of Skittles with 3 red, 4 purple, 3 green, 4 yellow, and 6 orange, what is the chance that I randomly pick and eat 3 orange Skittles in a row?
  • 6/20 * 5/19 * 4/18 = 0.01754

Class Survey Example

A class was surveyed to count how many students were wearing jeans. The results are shown in the table below.

• Probability that a randomly selected student wears jeans: 20/36 = 0.5556
• Probability that a female wears jeans: 8/19 = 0.4211
• Probability that a male wears jeans: 12/17 = 0.7059
• Probability that someone wearing jeans is male: 12/20 = 0.6
• Are being male and wearing jeans disjoint? No, 12 students fit both descriptors.
• Are being male and wearing jeans independent? No, P(jeans) = 0.5556, P(jeans | male) = 0.7059

Completing a Probability Table Example

• As of March 2017, roughly 20% of Americans were on Snapchat. The U.S. population is approximately 28% 13-34 year olds. Because Snapchat is popular with younger Americans, 17% of the country are 13-34 year olds who are on Snapchat. Construct a probability table.

• What is the probability that you use SnapChat if you are not 13-34 years old?
  • .03/.72 = 0.0417
• What is the probability that you are 13-34 years old if you use SnapChat?
  • .17/.20 = 0.85

AP Statistics Free-Response Question Example

3. A medical researcher surveyed a large group of men and women about whether they take medicine as prescribed.
   The responses were categorized as never, sometimes, or always. The relative frequency of each category is
   shown in the table.

(a) One person from those surveyed will be selected at random.

(i) What is the probability that the person selected will be someone whose response is never and who is a woman?

• P(never and woman) = 0.0636

(ii) What is the probability that the person selected will be someone whose response is never or who is a woman?

• P(never or woman) = P(never) + P(woman) - P(never and woman)
• = 0.12 + 0.53 - 0.0636
• = 0.5864

(iii) What is the probability that the person selected will be someone whose response is never given that the person is a woman?

P(never | woman) = \frac{P(never and woman)}{P(woman)} = \frac{0.0636}{0.53} = 0.12

(b) For the people surveyed, are the events of being a person whose response is never and being a woman independent? Justify your answer.

• Yes, the event of being a person who responds never is independent of the event of being a woman because
  P(never | woman) = P(never) = 0.12
• P(woman and never) = 0.0636 is the same as P(woman) x P(never) = (0.53)(0.12) = 0.0636.
• P(never | woman) = \frac{0.0636}{0.53} = 0.12 \text{ is the same as } P(never) = 0.12.
• P(woman | never) = \frac{0.0636}{0.12} = 0.53 \text{ is the same as } P(woman) = 0.53.

Common Mistakes

• Don’t use a simple probability rule when a general rule is appropriate.
• Don’t assume that two events are independent or disjoint without checking that they are.
• Don’t find probabilities for samples drawn without replacement as if they had been drawn with replacement.
• Don’t confuse “disjoint” with “independent.”

Key Concepts

• The probability rules from the previous chapter only work in special cases—when events are disjoint or independent.
• The General Addition Rule: P(A \cup B) = P(A) + P(B) - P(A \cap B)
• The General Multiplication Rule: P(A \cap B) = P(A) * P(B|A)
• Conditional Probability: P(B|A) = \frac{P(A \cap B)}{P(A)}

Independence

• We can use conditional probability to determine whether events are independent and to work with events that are not independent.
• Events A and B are independent if the occurrence of one does not affect the probability that the other occurs: P(B|A) = P(B).
• Events that are mutually exclusive (disjoint) cannot be independent—knowing that one happened means the other cannot happen.