Comprehensive Study Notes on Probability Terminology, Rules, and Events

Fundamental Terminology and Concepts of Probability

• Probability:

  • Defined as a measure associated with the degree of certainty regarding the outcomes of a particular experiment or activity.

  • Quantified as a number between $0$ and $1$, inclusive (that is, $0 \le P(A) \le 1$).

  • $P(A) = 0$ indicates that event $A$ can never happen.

  • $P(A) = 1$ indicates that event $A$ always happens.

  • $P(A) = 0.5$ indicates that event $A$ is equally likely to occur or not to occur.

  • The probability of any outcome represents the long-term relative frequency of that outcome.

• Experiment and Chance Experiment:

  • An experiment is a planned operation carried out under controlled conditions.

  • A chance experiment occurs when the result of the experiment is not predetermined.

  • Example: Flipping one fair coin twice.

• Outcome and Sample Space:

  • An outcome is a specific result of an experiment.

  • The sample space (denoted by the uppercase letter $S$) is the set of all possible outcomes of an experiment.

  • Three standard methods to represent a sample space are:

  • Listing all possible outcomes explicitly (e.g., flipping one fair coin yields $S = \{H, T\}$, where $H = \text{heads}$ and $T = \text{tails}$).

  • Constructing a tree diagram.

  • Constructing a Venn diagram.

• Events:

  • An event is any combination or subset of outcomes within a sample space.

  • Represented using uppercase letters such as $A$, $B$, $C$, etc.

  • The probability of event $A$ occurring is denoted $P(A)$.

• Equally Likely Outcomes:

  • Outcomes are considered equally likely if each outcome in the sample space has an identical probability of occurring.

  • Examples of equally likely outcomes:

  • Tossing a fair, six-sided die: each face ($1, 2, 3, 4, 5, 6$) is equally likely.

  • Tossing a fair coin: Head ($H$) and Tail ($T$) are equally likely.

  • Guessing randomly on a true/false exam question: selecting a correct answer or an incorrect answer is equally likely.

  • Calculation Formula for Equally Likely Outcomes:

  • $P(A) = \frac{\text{number of outcomes in event } A}{\text{total number of outcomes in sample space } S}$

  • Example: Tossing a fair dime and a fair nickel yields $S = \{HH, TH, HT, TT\}$ (total $4$ outcomes). Let $A = \text{getting one head}$ ($\{HT, TH\}$). Thus, $P(A) = \frac{2}{4} = 0.5$.

  • Example: Rolling one fair six-sided die with faces $\{1, 2, 3, 4, 5, 6\}$. Let $E = \text{rolling at least five}$ ($\{5, 6\}$). Thus, $P(E) = \frac{2}{6}$.

• Law of Large Numbers:

  • States that as the number of repetitions (trials) of an experiment increases, the empirical (observed) relative frequency of an event approaches its theoretical probability.

  • Although short-term outcomes do not follow a fixed pattern or order, the long-term relative frequency stabilizes near the theoretical value.

  • Empirical refers to observed experimental data.

• Fair vs. Biased Experiments:

  • Outcomes are not always equally likely; coins or dice can be unfair or biased.

  • European Euro Coin Study: Two European math professors had statistics students test the Belgian 1 Euro coin across $250$ trials, resulting in heads $56\%$ of the time and tails $44\%$ of the time, suggesting coin bias.

  • Standard vs. Casino Dice:

  • Standard home dice have small holes carved out and painted for spots, creating slight weight differences across faces that may cause bias.

  • Casino dice are manufactured with flat faces where holes are filled completely with paint of the exact same density as the die material to maintain strict fairness.

Operations on Events and Set Notation

• "OR" Event (Union):

  • An outcome is in the event $A \text{ OR } B$ if the outcome is in $A$, in $B$, or in both $A$ and $B$.

  • Represented using union notation: $A \cup B$.

  • Example: Let $A = \{1, 2, 3, 4, 5\}$ and $B = \{4, 5, 6, 7, 8\}$. Then $A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8\}$. Note that shared elements ($4$ and $5$) are listed only once.

• "AND" Event (Intersection):

  • An outcome is in the event $A \text{ AND } B$ if the outcome is in both $A$ and $B$ simultaneously.

  • Represented using intersection notation: $A \cap B$.

  • Example: Let $A = \{1, 2, 3, 4, 5\}$ and $B = \{4, 5, 6, 7, 8\}$. Then $A \cap B = \{4, 5\}$.

• Complement of an Event:

  • The complement of event $A$ is denoted $A'$ (read "A prime").

  • $A'$ consists of all outcomes in the sample space $S$ that are NOT in $A$.

  • Fundamental property of complements: $P(A) + P(A') = 1$.

  • Example: Let $S = \{1, 2, 3, 4, 5, 6\}$ and $A = \{1, 2, 3, 4\}$. Then $A' = \{5, 6\}$. Here $P(A) = \frac{4}{6}$, $P(A') = \frac{2}{6}$, and $P(A) + P(A') = \frac{4}{6} + \frac{2}{6} = 1$.

• Conditional Probability:

  • Written as $P(A|B)$ and read as "the probability of $A$ given $B$".

  • Represents the probability that event $A$ will occur given that event $B$ has already occurred.

  • A condition reduces the sample space from $S$ to $B$.

  • Formula:

  • $P(A|B) = \frac{P(A \cap B)}{P(B)}$, provided $P(B) > 0$

  • Example: Toss one fair six-sided die ($S = \{1, 2, 3, 4, 5, 6\}$). Let $A = \text{face is 2 or 3}$ ($\{2, 3\}$) and $B = \text{face is even}$ ($\{2, 4, 6\}$).

  • Using reduced sample space $B = \{2, 4, 6\}$: $1$ outcome ($2$) satisfies $A$, out of $3$ total outcomes in $B$. Thus, $P(A|B) = \frac{1}{3}$.

  • Using formula: $P(A \cap B) = \frac{1}{6}$, $P(B) = \frac{3}{6} \implies P(A|B) = \frac{1/6}{3/6} = \frac{1}{3}$.

Independent and Mutually Exclusive Events

• Independent Events:

  • Two events $A$ and $B$ are independent if knowing that one occurred does not alter the probability of the other occurring.

  • Two events $A$ and $B$ are independent if and only if at least one of the following equivalent conditions holds:

  • $P(A|B) = P(A)$

  • $P(B|A) = P(B)$

  • $P(A \cap B) = P(A)P(B)$

  • Showing any one of these three conditions is sufficient to prove independence.

  • If two events fail these conditions, they are dependent events.

  • Default rule: If it is unknown whether events are independent or dependent, assume they are dependent until proven otherwise.

• Sampling Methods and Independence:

  • Sampling With Replacement:

  • Each selected member of a population is replaced before the next draw.

  • Members can be chosen more than once.

  • Draws are independent events because population counts and probabilities remain constant across draws.

  • Sampling Without Replacement:

  • Each selected member is set aside and cannot be picked again.

  • Members can be chosen at most once.

  • Draws are dependent events because the total population size and available options change after each draw.

• Mutually Exclusive Events:

  • Events $A$ and $B$ are mutually exclusive if they cannot occur at the same time.

  • Mutually exclusive events share no outcomes in common: $A \cap B = \emptyset$.

  • Mathematical Definition:

  • $P(A \cap B) = 0$

  • Default rule: Assume events are NOT mutually exclusive until proven otherwise.

  • Critical Distinction: "Independent" and "Mutually Exclusive" are different concepts. Mutually exclusive events with non-zero probabilities can never be independent, because if one occurs, the chance of the other occurring drops to zero.

The Two Basic Rules of Probability

• The Multiplication Rule:

  • For any two events $A$ and $B$ defined on a sample space:

  • $P(A \cap B) = P(B)P(A|B)$

  • Equivalently: $P(A \cap B) = P(A)P(B|A)$

  • Rewritten for conditional probability:

  • $P(A|B) = \frac{P(A \cap B)}{P(B)}$

  • Special Case for Independent Events:

  • If $A$ and $B$ are independent, then $P(A|B) = P(A)$, simplifying the rule to:

  • $P(A \cap B) = P(A)P(B)$

• The Addition Rule:

  • For any two events $A$ and $B$ defined on a sample space:

  • $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

  • Special Case for Mutually Exclusive Events:

  • If $A$ and $B$ are mutually exclusive, then $P(A \cap B) = 0$, simplifying the rule to:

  • $P(A \cup B) = P(A) + P(B)$

Detailed Worked Examples and Practice Problems

• Example 3.1 & Try It 3.1 (Whole Number Sample Spaces):

  • Example 3.1:

  • Sample space $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19\}$ ($n(S) = 19$).

  • Let $A = \text{even numbers} = \{2, 4, 6, 8, 10, 12, 14, 16, 18\}$ ($n(A) = 9$).

  • Let $B = \text{numbers greater than 13} = \{14, 15, 16, 17, 18, 19\}$ ($n(B) = 6$).

  • Probabilities: $P(A) = \frac{9}{19}$, $P(B) = \frac{6}{19}$.

  • Intersections & Unions: $A \cap B = \{14, 16, 18\}$ ($n = 3$), $A \cup B = \{2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19\}$ ($n = 12$).

  • Values: $P(A \cap B) = \frac{3}{19}$, $P(A \cup B) = \frac{12}{19}$.

  • Complement: $A' = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}$, $P(A') = \frac{10}{19}$. Sum $P(A) + P(A') = \frac{9}{19} + \frac{10}{19} = 1$.

  • Conditionals: $P(A|B) = \frac{3}{6} = 0.5$; $P(B|A) = \frac{3}{9} = \frac{1}{3} \approx 0.3333$. Probabilities are not equal.

  • Try It 3.1:

  • $S = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}$ ($n(S) = 12$).

  • $A = \text{sum is even} = \{(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)\}$ ($n(A) = 6$).

  • $B = \text{first number is prime } (2, 3) = \{(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}$ ($n(B) = 8$).

  • Probabilities: $P(A) = \frac{6}{12} = 0.5$, $P(B) = \frac{8}{12} = \frac{2}{3}$.

  • $A \cap B = \{(2,2), (2,4), (3,1), (3,3)\}$ ($n = 4$) $\implies P(A \cap B) = \frac{4}{12} = \frac{1}{3}$.

  • $A \cup B$ has $10$ elements $\implies P(A \cup B) = \frac{10}{12} = \frac{5}{6}$.

  • Complement $B' = \{(1,1), (1,2), (1,3), (1,4)\}$ ($n = 4$) $\implies P(B') = \frac{4}{12} = \frac{1}{3}$.

  • Conditionals: $P(A|B) = \frac{4}{8} = 0.5$; $P(B|A) = \frac{4}{6} = \frac{2}{3}$. Not equal.

• Example 3.2 & Try It 3.2 (Die Rolls & Integer Selection):

  • Example 3.2 (Fair Six-Sided Die):

  • $S = \{1, 2, 3, 4, 5, 6\}$.

  • $T = \{2\} \implies P(T) = \frac{1}{6}$.

  • A = \text{even} = \{2, 4, 6\} \implies P(A) = \frac{3}{6} = 0.5$.\n - B = \text{less than four} = {1, 2, 3} \implies P(B) = \frac{3}{6} = 0.5$.

  • Complement A' = \{1, 3, 5\} \implies P(A') = 0.5$.\n - Conditionals: P(A|B) = \frac{1}{3}$,$P(B|A) = \frac{1}{3}.\n - A \cap B = {2} \implies P(A \cap B) = \frac{1}{6}$.

  • A \cup B = \{1, 2, 3, 4, 6\} \implies P(A \cup B) = \frac{5}{6}$.\n - A \cup B' = {2, 4, 5, 6} \implies P(A \cup B') = \frac{4}{6} = \frac{2}{3}$.

  • N = \text{prime} = \{2, 3, 5\} \implies P(N) = \frac{3}{6} = 0.5$.\n - I = \text{seven} = \emptyset \implies P(I) = 0$.

  • Try It 3.2 (Selecting Number 1 to 10):

  • $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.

  • F = \{5\} \implies P(F) = 0.1$.\n - A = \text{more than 6} = {7, 8, 9, 10} \implies P(A) = 0.4$.

  • B = \text{odd} = \{1, 3, 5, 7, 9\} \implies P(B) = 0.5$.\n - B' = {2, 4, 6, 8, 10} \implies P(B') = 0.5$.

  • $A|B = \{7, 9\}$ out of B \implies P(A|B) = \frac{2}{5} = 0.4$.\n - B|A = {7, 9}$out of$A \implies P(B|A) = \frac{2}{4} = 0.5$.

  • A \cup B = \{1, 3, 5, 7, 8, 9, 10\} \implies P(A \cup B) = 0.7$.\n - A' \cup B = {1, 2, 3, 4, 5, 6, 7, 9} \implies P(A' \cup B) = 0.8$.

  • P = \text{composite} = \{4, 6, 8, 9, 10\} \implies P(P) = 0.5$.\n - M = \text{multiple of 3} = {3, 6, 9} \implies P(M) = 0.3$.

• Example 3.3 & Try It 3.3 (Contingency Table Analysis):

  • Example 3.3 (Handedness and Sex at Birth, Total $N = 100$):

  • Data Table:

    • Males ($M$): Right-handed ($R$) = $43$, Left-handed ($L$) = $9$ (Total $M = 52$).

    • Females ($F$): Right-handed ($R$) = $44$, Left-handed ($L$) = $4$ (Total $F = 48$).

    • Column Totals: $R = 87$, $L = 13$.

  • Results:

    • $P(M) = \frac{52}{100} = 0.52$

    • $P(F) = \frac{48}{100} = 0.48$

    • $P(R) = \frac{87}{100} = 0.87$

    • $P(L) = \frac{13}{100} = 0.13$

    • $P(M \cap R) = \frac{43}{100} = 0.43$

    • $P(F \cap L) = \frac{4}{100} = 0.04$

    • $P(M \cup F) = \frac{100}{100} = 1.0$

    • $P(M \cup R) = 0.52 + 0.87 - 0.43 = 0.96$

    • $P(F \cup L) = 0.48 + 0.13 - 0.04 = 0.57$

    • $P(M') = 1 - 0.52 = 0.48$

    • $P(R|M) = \frac{43}{52} \approx 0.8269$

    • $P(F|L) = \frac{4}{13} \approx 0.3077$

    • $P(L|F) = \frac{4}{48} = \frac{1}{12} \approx 0.0833$

  • Try It 3.3 (Beverage Preference and Gender, Total $N = 100$):

  • Data Table:

    • Men ($M$): Tea ($T$) = $22$, Coffee ($C$) = $26$ (Total $M = 48$).

    • Women ($W$): Tea ($T$) = $16$, Coffee ($C$) = $36$ (Total $W = 52$).

    • Column Totals: $T = 38$, $C = 62$.

  • Results:

    • $P(M) = 0.48$, $P(W) = 0.52$, $P(T) = 0.38$, $P(C) = 0.62$

    • $P(M \cap T) = 0.22$, $P(W \cap C) = 0.36$

    • $P(M \cup W) = 1.0$

    • $P(M \cup T) = 0.48 + 0.38 - 0.22 = 0.64$

    • $P(W \cup C) = 0.52 + 0.62 - 0.36 = 0.78$

    • $P(M') = 1 - 0.48 = 0.52$

    • $P(T|M) = \frac{22}{48} \approx 0.4583$

    • $P(W|C) = \frac{36}{62} \approx 0.5806$

    • $P(C|W) = \frac{36}{52} \approx 0.6923$

• Examples 3.4 & 3.5 / Try Its 3.4 & 3.5 (Card Sampling Protocols):

  • Deck structure: $52$ total cards, $4$ suits (Clubs $C$, Diamonds $D$, Hearts $H$, Spades $S$), $13$ ranks per suit ($1$ to $10$, $J$, $Q$, $K$).

  • Example 3.4:

  • With replacement: Sequence $\{Q\text{ of spades}, 10\text{ of clubs}, Q\text{ of spades}\}$ allows duplicate $Q\text{ of spades}$.

  • Without replacement: Sequence $\{K\text{ of hearts}, 3\text{ of diamonds}, J\text{ of spades}\}$ contains no duplicates.

  • Try It 3.4:

  • Sequence $\{Q\text{ of spades}, K\text{ of hearts}, Q\text{ of spades}\}$ must be with replacement due to duplicate card.

  • Sequence $\{Q\text{ of spades}, K\text{ of hearts}, J\text{ of spades}\}$ could occur under either sampling method.

  • Example 3.5:

  • Sequence $QS, 1D, 1C, QD$ without replacing cards = sampling without replacement.

  • Sequence $KH, 7D, 6D, KH$ replacing cards each time = sampling with replacement.

  • Try It 3.5:

  • Outcome $QS, 1D, 1C, QD$: Possible both with and without replacement.

  • Outcome $KH, 7D, 6D, KH$: Possible only with replacement ($KH$ repeated).

  • Outcome $QS, 7D, 6D, KS$: Possible both with and without replacement.

• Examples 3.6 & 3.7 / Try Its 3.6 & 3.7 (Coin Flips, Cards, and Balls):

  • Example 3.6 (Flipping Two Fair Coins):

  • $S = \{HH, HT, TH, TT\}$.

  • $A = \text{at most 1 tail} = \{HH, HT, TH\} \implies P(A) = \frac{3}{4}$.

  • $B = \text{all tails} = \{TT\} \implies P(B) = \frac{1}{4}$. Here $B = A'$.

  • $C = \text{all heads} = \{HH\} \implies P(B \cap C) = 0$ ($B, C$ mutually exclusive).

  • D = \text{more than 1 tail} = \{TT\} \implies P(D) = \frac{1}{4}$.\n - E = \text{head on first flip} = {HT, HH} \implies P(E) = \frac{2}{4} = 0.5$.

  • F = \text{at least 1 tail} = \{HT, TH, TT\} \implies P(F) = \frac{3}{4}$.\n - **Try It 3.6 (Two Cards With Replacement)**:\n - Find probability of at least one black card.\n - P(\text{both red}) = 0.5 \times 0.5 = 0.25$.

  • P(\text{at least 1 black}) = 1 - P(\text{both red}) = 1 - 0.25 = 0.75$.\n - **Example 3.7**:\n - F = \text{at most 1 tail} = {HH, HT, TH} \implies P(F) = 0.75$.

  • G = \text{two same faces} = \{HH, TT\} \implies P(G) = 0.5$.\n - H = \text{head on first flip} = {HH, HT} \implies P(H) = 0.5$.

  • $F \cap G = \{HH\} \implies P(F \cap G) = 0.25 \ne 0 \implies F, G$ are not mutually exclusive.

  • $J = \text{all tails} = \{TT\} \implies J \cap H = \emptyset \implies P(J \cap H) = 0 \implies J, H$ are mutually exclusive.

  • Try It 3.7 (Box with 1 White, 1 Red Ball, Sampling With Replacement):

  • $S = \{WW, WR, RW, RR\}$.

  • F = \text{white twice} = \{WW\} \implies P(F) = 0.25$.\n - G = \text{different colors} = {WR, RW} \implies P(G) = 0.5$.

  • H = \text{white on first pick} = \{WW, WR\} \implies P(H) = 0.5$.\n - F \cap G = \emptyset \implies P(F \cap G) = 0 \implies F, G are mutually exclusive.\n - G \cap H = {WR} \implies P(G \cap H) = 0.25 e 0 \implies G, H are not mutually exclusive.\n\n- **Examples 3.8 & 3.9 / Try Its 3.8 & 3.9 (Die, Language, Math/Science, Marbles)**:\n - **Example 3.8 (Single Die Roll)**:\n - S = {1, 2, 3, 4, 5, 6}$,$A = {1, 3, 5}$,$B = {2, 4, 6}.\n - C = \text{odd } > 2 = {3, 5}$,$D = \text{even } < 5 = {2, 4}$.$P(C \cap D) = 0 \implies C, D mutually exclusive.\n - E = \text{faces } < 5 = {1, 2, 3, 4}$. Are$C$and$E$mutually exclusive? No, because$C \cap E = {3, 5} \implies P(C \cap E) = \frac{2}{6} e 0$.

  • $P(C|A) = \frac{2}{3}$.

  • Try It 3.8 (Language Learning):

  • $P(A) = 0.4$, $P(B) = 0.2$, P(A \cap B) = 0.08$.\n - Test independence: P(A)P(B) = 0.4 \times 0.2 = 0.08 = P(A \cap B)$. Events$A$and$B are independent.\n - **Example 3.9 (Math and Science Classes)**:\n - P(G) = 0.6$,$P(H) = 0.5$,$P(G \cap H) = 0.3$.

  • P(G|H) = \frac{P(G \cap H)}{P(H)} = \frac{0.3}{0.5} = 0.6 = P(G)$.\n - P(G)P(H) = 0.6 \times 0.5 = 0.3 = P(G \cap H)$.

  • Thus, $G$ and $H$ are independent.

  • Try It 3.9 (Red and Green Marbles in a Bag):

  • Bag contains $6$ red marbles ($R1, R2, R3, R4, R5, R6$) and $4$ green marbles ($G1, G2, G3, G4$). Total N = 10$.\n - Sample space S = {R1, R2, R3, R4, R5, R6, G1, G2, G3, G4}.\n - Event G = \text{green marble}$, Event$O = \text{odd-numbered marble}.\n - Outcomes in G \cap O = {G1, G3}.\n - P(G \cap O) = \frac{2}{10} = 0.2$.

• Example 3.10 & Try It 3.10 (Class Enrolments & Library Checks):

  • Example 3.10:

  • $P(C) = 0.75$, $P(D) = 0.3$, $P(C|D) = 0.75$, P(C \cap D) = 0.225$.\n - a. Independent? Yes, because P(C|D) = 0.75 = P(C)$(and$P(C)P(D) = 0.75 \times 0.3 = 0.225 = P(C \cap D)).\n - b. Mutually exclusive? No, because P(C \cap D) = 0.225 e 0$.

  • c. P(D|C) = \frac{P(C \cap D)}{P(C)} = \frac{0.225}{0.75} = 0.3$.\n - **Try It 3.10**:\n - P(B) = 0.40$,$P(D) = 0.30$,$P(B \cap D) = 0.20$.

  • a. P(B|D) = \frac{0.20}{0.30} = \frac{2}{3} \approx 0.6667$.\n - b. P(D|B) = \frac{0.20}{0.40} = 0.50$.

  • c. Independent? No, because $P(B|D) = \frac{2}{3} \ne P(B) = 0.40$.

  • d. Mutually exclusive? No, because P(B \cap D) = 0.20 \ne 0$.\n\n- **Example 3.11 & Try It 3.11 (Numbered Cards & Arena Fans)**:\n - **Example 3.11**:\n - Box with 3$red cards ($R1, R2, R3$) and$5$blue cards ($B1, B2, B3, B4, B5$). Total$N = 8$.

  • $P(R) = \frac{3}{8}$, $P(B) = \frac{5}{8}$, $P(R \cap B) = 0$ (mutually exclusive).

  • Even cards E = \{R2, B2, B4\} \implies P(E) = \frac{3}{8}$.\n - Conditionals: P(E|B) = \frac{2}{5}$,$P(B|E) = \frac{2}{3}$.

  • Let G = \text{cards } > 3 = \{B4, B5\} \implies P(G) = \frac{2}{8} = \frac{1}{4}$.\n - Let H = \text{blue cards between 1 and 4} = {B1, B2, B3, B4}$.

  • $P(G|H) = \frac{1}{4}$. Since $P(G) = P(G|H) = \frac{1}{4}$, $G$ and $H$ are independent.

  • Try It 3.11:

  • P(\text{home}) = 0.70 \implies P(\text{away}) = P(A) = 0.30$.\n - P(\text{blue}) = P(B) = 0.25$.

  • Given: $P(B \cap A) = 0.20$, P(B|A) = 0.67$.\n - Independent? No, because P(B|A) = 0.67 e P(B) = 0.25$.

  • Mutually exclusive? No, because P(B \cap A) = 0.20 \ne 0$.\n\n- **Example 3.12 & Try It 3.12 (Class Hair Length & Route Selection)**:\n - **Example 3.12**:\n - Given: P(W) = 0.60$,$P(L) = 0.50$,$P(W \cap L) = 0.45$,$P(L|W) = 0.75$.

  • Independent? No, because P(L|W) = 0.75 \ne P(L) = 0.50$.\n - **Try It 3.12**:\n - Routes I$(Interstate) and$F$(Fifth Street).$P(I) = 0.44$,$P(F) = 0.56$,$P(I \cap F) = 0$.

  • P(I \cup F) = P(I) + P(F) - P(I \cap F) = 0.44 + 0.56 - 0 = 1.00$.\n\n- **Example 3.13 & Try It 3.13 (Coin/Die Combination & Ball Draw)**:\n - **Example 3.13**:\n - Coin outcomes = 2$($H, T$); Die outcomes =$6$($1, 2, 3, 4, 5, 6).\n - Total sample space size = 2 \times 6 = 12.\n - Outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.\n - A = \text{heads followed by even} = {H2, H4, H6} \implies P(A) = \frac{3}{12} = 0.25$.

  • B = \text{heads followed by 3} = \{H3\} \implies P(B) = \frac{1}{12}$.\n - Mutually exclusive? Yes, P(A \cap B) = 0$.

  • Independent? No, P(A)P(B) = \frac{1}{4} \times \frac{1}{12} = \frac{1}{48} \ne P(A \cap B) = 0$.\n - **Try It 3.13**:\n - Box with 1$white,$1$red ball (sampling with replacement,$2 draws).\n - T = \text{white twice} = {WW}$, F = \text{white first} = \{WW, WR\}$, S = \text{white second} = {WW, RW}$.

  • a. P(T) = \frac{1}{4} = 0.25$.\n - b. P(T|F) = \frac{1/4}{2/4} = 0.50$.

  • c. $T$ and $F$ independent? No, P(T|F) = 0.50 \ne P(T) = 0.25$.\n - d. F$and$S$mutually exclusive? No,$F \cap S = {WW} \implies P(F \cap S) = 0.25 e 0$.

  • e. $F$ and $S$ independent? Yes, $P(F)P(S) = 0.5 \times 0.5 = 0.25 = P(F \cap S)$.

• Example 3.14 & Try It 3.14 (Vacations and Car Purchases):

  • Example 3.14:

  • $A = \text{New Zealand}$, $B = \text{Alaska}$. $P(A) = 0.60$, $P(B) = 0.35$, P(A \cap B) = 0$.\n - P(A \cup B) = 0.60 + 0.35 = 0.95$.

  • Probability of no vacation = 1 - 0.95 = 0.05$.\n - **Try It 3.14**:\n - P(A) = 0.25$,$P(B) = 0.65$,$P(A \cap B) = 0$.

  • P(A \cap B) = 0$.\n - P(A \cup B) = 0.25 + 0.65 = 0.90$.

• Example 3.15 & Try It 3.15 (Sports Performance Probabilities):

  • Example 3.15 (Soccer Goals):

  • $P(A) = 0.65$, $P(B) = 0.65$, P(B|A) = 0.90$.\n - a. P(A \cap B) = P(A)P(B|A) = 0.65 \times 0.90 = 0.585$.

  • b. P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.65 + 0.65 - 0.585 = 0.715$.\n - c. Independent? No, P(B|A) = 0.90 e P(B) = 0.65$.

  • d. Mutually exclusive? No, P(A \cap B) = 0.585 \ne 0$.\n - **Try It 3.15 (Basketball Free Throws)**:\n - P(C) = 0.75$,$P(D) = 0.75$,$P(D|C) = 0.85$.

  • P(\text{makes both}) = P(C \cap D) = P(C)P(D|C) = 0.75 \times 0.85 = 0.6375$.\n\n- **Example 3.16 & Try It 3.16 (Swim Team and High School Senior Counts)**:\n - **Example 3.16 (Swim Team, Total N = 150)**:\n - Groups: Advanced (A = 75$), Intermediate ($I = 47$), Novice ($N = 150 - 75 - 47 = 28).\n - Practice 4x/week (P$): Advanced =$40$, Intermediate =$30$, Novice =$10$. Total$P = 80$.

  • a. P(N) = \frac{28}{150} \approx 0.1867$.\n - b. P(P) = \frac{80}{150} = \frac{8}{15} \approx 0.5333$.

  • c. P(A \cap P) = \frac{40}{150} = \frac{4}{15} \approx 0.2667$.\n - d. P(A \cap I) = 0. Mutually exclusive because a swimmer cannot belong to two skill levels simultaneously.\n - e. Independent? P(N) = \frac{28}{150} \approx 0.1867$vs$P(N|P) = \frac{10}{80} = 0.125$. Since$P(N|P) e P(N), they are dependent.\n - **Try It 3.16 (High School Seniors, Total N = 200)**:\n - Destinations: College = 140$, Work =$40$, Gap Year =$200 - 140 - 40 = 20.\n - Sports participants: College sports = 50$, Work sports =$30$, Gap Year sports =$5$. Total sports =$85$.

  • Probability senior is taking a gap year: P(\text{Gap Year}) = \frac{20}{200} = 0.10$.\n\n- **Example 3.17 & Try It 3.17 (Course Enrollments & Library Items)**:\n - **Example 3.17 (Felicity at Modesto JC)**:\n - P(M) = 0.20$,$P(S) = 0.65$,$P(M|S) = 0.25$.

  • a. P(M \cap S) = P(M|S)P(S) = 0.25 \times 0.65 = 0.1625$.\n - b. P(M \cup S) = P(M) + P(S) - P(M \cap S) = 0.20 + 0.65 - 0.1625 = 0.6875$.

  • c. Independent? No, P(M|S) = 0.25 \ne P(M) = 0.20$.\n - d. Mutually exclusive? No, P(M \cap S) = 0.1625 e 0$.

  • Try It 3.17 (Library Book and DVD Borrowing):

  • $P(B) = 0.40$, $P(D) = 0.30$, P(D|B) = 0.50$.\n - a. P(B \cap D) = P(D|B)P(B) = 0.50 \times 0.40 = 0.20$.

  • b. P(B \cup D) = P(B) + P(D) - P(B \cap D) = 0.40 + 0.30 - 0.20 = 0.50$.\n\n- **Example 3.18, 3.19 & Try Its 3.18, 3.19 (Medical Screening & Senior Probabilities)**:\n - **Example 3.18 (Breast Cancer Screening)**:\n - Given parameters: P(B) = \frac{1}{7} \approx 0.1429$,$P(N|B) = 0.02$,$P(N) = 0.85$.

  • a. $P(B) = \frac{1}{7} \approx 0.1429$, P(N) = 0.85$.\n - b. P(N|B) = 0.02$.

  • c. P(B \cap N) = P(N|B)P(B) = 0.02 \times \frac{1}{7} \approx 0.002857$.\n - d. P(B \cup N) = P(B) + P(N) - P(B \cap N) = \frac{1}{7} + 0.85 - 0.002857 \approx 0.9900$.

  • e. Independent? No, P(N|B) = 0.02 \ne P(N) = 0.85$.\n - f. Mutually exclusive? No, P(B \cap N) \approx 0.002857 e 0$.

  • Try It 3.18 (High School Senior College and Sports Probability):

  • Total seniors = $200$, Seniors going to college and playing sports = 50$.\n - P(\text{College } \cap \text{ Sports}) = \frac{50}{200} = 0.25$.

  • Example 3.19 (Follow-up to Example 3.18, $P = \text{tests positive}$):

  • a. P(P|B) = 1 - P(N|B) = 1 - 0.02 = 0.98$.\n - b. P(B \cap P) = P(P|B)P(B) = 0.98 \times \frac{1}{7} \approx 0.1400$.

  • c. P(B') = 1 - P(B) = 1 - \frac{1}{7} = \frac{6}{7} \approx 0.8571$.\n - d. P(P) = 1 - P(N) = 1 - 0.85 = 0.15$.

  • Try It 3.19 (Library Complementary and Joint Probabilities):

  • Given: $P(B) = 0.40$, $P(D) = 0.30$, P(D|B) = 0.50$.\n - a. P(B') = 1 - P(B) = 1 - 0.40 = 0.60$.

  • b. P(D \cap B) = P(D|B)P(B) = 0.50 \times 0.40 = 0.20$.\n - c. P(B|D) = \frac{P(D \cap B)}{P(D)} = \frac{0.20}{0.30} = \frac{2}{3} \approx 0.6667$.

  • d. P(D \cap B') = P(D) - P(D \cap B) = 0.30 - 0.20 = 0.10$.\n - e. P(D|B') = \frac{P(D \cap B')}{P(B')} = \frac{0.10}{0.60} = \frac{1}{6} \approx 0.1667$.