Honors Geometry Unit 12 Review - Probability exhaustive Study Notes

Fundamental Probability Definitions

• Simple Event: This is defined as an event characterized by having only a single outcome.

• Not Simple Event / Compound Event: This is defined as a combination of two or more simple events.

• Empirical Probability: This represents the ratio of the number of outcomes in which a specified event occurs to the total number of trials conducted.

• Classical Probability: This is the statistical concept used to measure the likelihood of a specific event occurring.

• Mutually Exclusive: This term describes two or more events that have no outcomes in common; they cannot occur at the same time.

• Not Mutually Exclusive: This describes two or more events that share outcomes in common.

• Independent Event: Two events are independent if the occurrence of each event does not affect the occurrence or outcome of the other.

• Dependent Event: Two events are dependent if the occurrence of each event does affect the outcome of the other.

Probability Formulas

• Basic Probability:     $\frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$

• Conditional Probability:     $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$

• Multiplication Rule (for Independent Events):     $P(A \text{ and } B) = P(A) \times P(B)$

• Multiplication Rule (for Dependent Events):     $P(A \text{ and } B) = P(A) \times P(B|A)$

• Addition Rule (for Mutually Exclusive Events):     $P(A \text{ or } B) = P(A) + P(B)$

• Addition Rule (for events that are Not Mutually Exclusive):     $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

• Permutation of $n$ events:     $n!$

• Permutation of $n$ events taken $r$ at a time:     ${}_n P_r = \frac{n!}{(n-r)!}$

• Distinguishable Permutations:     $\frac{n!}{n_1! n_2! \dots}$

• Combinations:     ${}_n C_r = \frac{n!}{r! (n-r)!}$

Sample Space and Basic Probability Scenarios

• Milky Way Candy Choices: Identifying the sample space for choices between sizes (Fun-sized, King-size, or Regular-size) and varieties (Regular or Midnight).     * Sample Space: $\text{{Fun Reg, Fun Midnight, King Reg, King Midnight, Reg Reg, Reg Midnight}}$

• Coin Flipping: Identifying the sample space for flipping 3 coins.     * Sample Space: $\text{{hhh, hht, hth, htt, thh, tht, tth, ttt}}$

• Spinner Probability: Given a 7-color spinner, the probability of spinning a primary color (Red, Yellow, Blue):     * $P = \frac{3}{7}$

• Survey Data: In a survey of college students, $880$ cheated and $1721$ did not. Probability of selecting a student who cheated:     * $\text{Total} = 880 + 1721 = 2601$     * $P(\text{Cheated}) = \frac{880}{2601}$

• Event Classification Examples:     * Case 1: Event A: A red skittle is selected and eaten. Event B: A blue candy is then selected from the same package and eaten.         * Classification: Dependent.     * Case 2: A die is rolled. Event A: Even number. Event B: Number greater than $3$.         * Classification: Not Mutually Exclusive (they share the numbers $4$ and $6$).     * Case 3: Event A: Student taking Science. Event B: Student taking Calculus.         * Classification: Mutually Exclusive (per document context).

Card Drawing Probability Scenarios

• Two Consecutive Red Cards (Without Replacement):     * $P = \frac{26}{52} \times \frac{25}{51} = \frac{650}{2652} = \frac{25}{102}$

• Two Consecutive Red Cards (With Replacement):     * $P = \frac{26}{52} \times \frac{26}{52} = \frac{676}{2704} = \frac{1}{4}$

• Drawing a 5 OR a 6:     * $P = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$

• Drawing an Ace OR a Black Card:     * $P = P(\text{Ace}) + P(\text{Black}) - P(\text{Ace and Black})$     * $P = \frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}$

• Three Consecutive Aces (Without Replacement):     * $P = \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} = \frac{24}{132600} = \frac{1}{5525}$

• Three Consecutive Aces (With Replacement):     * $P = \frac{4}{52} \times \frac{4}{52} \times \frac{4}{52} = \frac{64}{140608} = \frac{1}{2197}$

Categorical and Conditional Probability Distributions

Students Traveling to Basketball Game in Columbia

• Probability student is female: $\frac{245}{507}$

• Probability student is driving: $\frac{347}{507}$

• Probability student is taking the bus AND is male: $\frac{78}{507}$

• Probability student is taking the bus OR is male:     * $(78 + 82 + 184) / 507 = \frac{344}{507}$

• Probability a female student is driving: $\frac{163}{245}$

• Probability a student who is driving is male: $\frac{184}{347}$

Blood Type Distribution (Sample size 100 Americans)

• Outcome: The probability of NOT selecting a person with blood type B+:     * $\text{B+ count} = 10$     * $\text{Not B+} = 100 - 10 = 90$     * $P(\text{Not B+}) = \frac{90}{100} = \frac{9}{10}$

Student Scent Qualitative Analysis

• P(man): $\frac{192}{405} = \frac{64}{135}$

• P(man and smells bad): $\frac{52}{405}$

• P(man or smells bad): $\frac{135 + 52 + 5 + 21}{405} = \frac{213}{405}$

• P(man given smells bad): $\frac{52}{73}$

• P(smells bad and no smell): This was calculated as the union of independent counts $\frac{73+10}{405} = \frac{83}{405}$. (Note: There is no intersection between 'Smells Bad' and 'No Smell' columns; calculated as a combined probability as per transcript).

• P(smells bad given no smell): $\frac{0}{10} = 0$ (Note: Manual calculation in text varies, value $\frac{10}{405}$ is noted but standard math for 'given' suggests $\frac{0}{10}$ if events are disjoint).

• P(smells bad or smells good): $\frac{322 + 73}{405} = \frac{395}{405} = \frac{79}{81}$

Boston Red Sox 2007 Win Probabilities

• Context: Win percentage = $59.3\% = 0.593$. Outcomes are treated as independent events.

• Win 2 in a row: $(0.593) \times (0.593) = 0.352$

• Lose 2 in a row: Probability of loss = $1 - 0.593 = 0.407$.     * $P = (0.407) \times (0.407) = 0.166$

• Win 7 in a row: $(0.593)^7 = 0.026$

• Lose at least one of next 7 games: $1 - P(\text{Win all 7}) = 1 - 0.026 = 0.974$

Counting Principles: Permutations and Combinations

• Employee Clock-in Codes (4 digits, 0-9):     * Scenario A: First number cannot be 0, repetitions allowed.         * $9 \times 10 \times 10 \times 10 = 9,000$     * Scenario B: First number cannot be 0 or 1, others must be 1-6 and no repetition.         * $8 \times 6 \times 5 \times 4 = 960$

• McDonald's Lunch Tree Diagram:     * Choices: Main (Big Mac, Filet-o-Fish, Chicken Selects).     * Sandwich sides: French Fries or Apple Dippers.     * Chicken Selects sides: BBQ, Ranch, or Sweet-n-Sour.     * Drinks: Coke, Sprite, or Dr. Pepper.     * Probabilities (based on 21 total paths):         * $P(\text{French Fries}) = \frac{6}{21}$         * $P(\text{Filet-o-Fish and Coke}) = \frac{2}{21}$         * $P(\text{Sandwich and French Fries}) = \frac{6}{21}$         * $P(\text{Chicken Selects given Ranch}) = \frac{3}{3} = 1$ (given path definition).         * $P(\text{BBQ given Chicken Selects}) = \frac{3}{9} = \frac{1}{3}$         * $P(\text{Chicken Selects or Big Mac}) = \frac{9 + 6}{21} = \frac{15}{21}$

• Race Standings: 9 swimmers in a race, assuming no ties. How many standings?     * ${}_9 P_9 = 9! = 362,880$

• Golf Club Bags: Choosing 14 clubs out of 26 available.     * ${}<em>{26} C</em>{14} = 9,657,700$

• Distinguishable Permutations of "HANNAH":     * Letters: H=2, A=2, N=2 (Total = 6).     * $\frac{6!}{2! 2! 2!} = \frac{720}{8} = 90$

• Seating Arrangements: 4 students entering a room with 10 vacant seats.     * ${}_{10} P_4 = 5040$

• Simple Random Samples: Population size 90, sample size 6.     * ${}_{90} C_6 = 622,614,630$