Probability, Conditional Probability, and Expected Value Study Guide

Basic Concepts of Probability

• Union Rule for Probability:   For any two events $E$ and $F$, the probability of their union (the probability that event $E$ or event $F$ or both occur) is calculated as:   $P(E \cup F) = P(E) + P(F) - P(E \cap F)$

• Union Rule for Mutually Exclusive Events:   Two events $E$ and $F$ are defined as mutually exclusive if they cannot occur at the same time, meaning their intersection is empty ($E \cap F = \emptyset$ and $P(E \cap F) = 0$). For mutually exclusive events, the union rule simplifies to:   $P(E \cup F) = P(E) + P(F)$

• Example 1: Card Selection   Draw a single card from a standard deck of $52$ playing cards. Find the probability that the card drawn is a heart or a king.

  • Let $H$ be the event of drawing a heart: $P(H) = \frac{13}{52}$.

  • Let $K$ be the event of drawing a king: $P(K) = \frac{4}{52}$.

  • The event of drawing a card that is both a heart and a king is $H \cap K$ (the King of Hearts): $P(H \cap K) = \frac{1}{52}$.

  • Applying the Union Rule:     $P(H \cup K) = P(H) + P(K) - P(H \cap K) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \approx 0.3077$

• Example 2: Two Dice Outcomes

  Roll two fair six-sided dice. The total number of equally likely outcomes in the sample space is $6 \times 6 = 36$.

• Part a: Find the probability that the first die shows a $3$ or $4$, or the sum of the two dice is $8$.

  • Let $E$ = First die is a $3$ or $4$. The outcomes are:       $E = \{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)\}$       $n(E) = 12 \implies P(E) = \frac{12}{36}$

  • Let $F$ = The sum of the dice is $8$. The outcomes are:       $F = \{(2,6), (3,5), (4,4), (5,3), (6,2)\}$       $n(F) = 5 \implies P(F) = \frac{5}{36}$

  • The intersection $E \cap F$ contains outcomes where the first die is $3$ or $4$ AND the sum is $8$:       $E \cap F = \{(3,5), (4,4)\}$       $n(E \cap F) = 2 \implies P(E \cap F) = \frac{2}{36}$

  • Using the Union Rule:       $P(E \cup F) = P(E) + P(F) - P(E \cap F) = \frac{12}{36} + \frac{5}{36} - \frac{2}{36} = \frac{15}{36} = \frac{5}{12} \approx 0.4167$

• Part b: Find the probability that the sum is $11$ or exactly one die shows a $3$.

  • Let $E$ = The sum is $11$: $E = \{(5,6), (6,5)\} \implies P(E) = \frac{2}{36}$.

  • Let $F$ = Exactly one die shows a $3$. Outcomes are:       $F = \{(3,1), (3,2), (3,4), (3,5), (3,6), (1,3), (2,3), (4,3), (5,3), (6,3)\}$       Note that $(3,3)$ is excluded because both dice show a $3$. Thus, $n(F) = 10 \implies P(F) = \frac{10}{36}$.

  • Check intersection $E \cap F$: If a die shows a $3$, the maximum possible sum is $3 + 6 = 9$. A sum of $11$ cannot contain a $3$. Thus, $E \cap F = \emptyset$ and $P(E \cap F) = 0$ (mutually exclusive).

  • Applying the Union Rule:       $P(E \cup F) = P(E) + P(F) = \frac{2}{36} + \frac{10}{36} = \frac{12}{36} = \frac{1}{3} \approx 0.3333$

  • Complement Rule:   The complement of an event $E$, denoted as $E'$, consists of all outcomes in the sample space that are not in $E$.

• Formula for probability of event $E$:     $P(E) = 1 - P(E')$

• Formula for probability of complement $E'$:     $P(E') = 1 - P(E)$

  • Example 3: Complement Rule Applications

• Part a: Roll two dice. Find the probability that a sum of $7$ is not rolled.

  • Let $E$ be the event that a sum of $7$ is rolled: $E = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$.

  • $P(E) = \frac{6}{36} = \frac{1}{6}$.

  • The event that a sum of $7$ is NOT rolled is $E'$. Applying the complement rule:       $P(E') = 1 - P(E) = 1 - \frac{1}{6} = \frac{5}{6} \approx 0.8333$

• Part b: There is a $30\%$ chance of rain tomorrow ($P(\text{rain}) = 0.30$). Determine the probability that it will not rain tomorrow.

  • $P(\text{no rain}) = 1 - P(\text{rain}) = 1 - 0.30 = 0.70 = 70\%$

  • Odds:

• If the odds favoring event $E$ are $m$ to $n$ (written $m:n$), then:     $P(E) = \frac{m}{m + n}$     $P(E') = \frac{n}{m + n}$

• If the odds favoring event $E$ are $m$ to $n$, then the odds against event $E$ are $n$ to $m$.

• The odds in favor of event $E$ represent the ratio of the probability that $E$ occurs to the probability that $E$ does not occur:     $\text{Odds in favor of } E = \frac{P(E)}{P(E')}, \quad P(E') \neq 0$

  • Example 4: Odds Calculations

• Part a: The probability of rain tomorrow is $0.30$. Find the odds in favor of rain tomorrow.

  • $P(\text{rain}) = 0.30$, $P(\text{no rain}) = 1 - 0.30 = 0.70$

  • $\text{Odds in favor} = \frac{0.30}{0.70} = \frac{3}{7} \implies 3 \text{ to } 7 \text{ (or } 3:7)$

• Part b: Suppose the odds in favor of passing an exam are $7$ to $2$. Find the probability of passing the exam.

  • Given $m = 7$ and $n = 2$:       $P(\text{passing}) = \frac{m}{m + n} = \frac{7}{7 + 2} = \frac{7}{9} \approx 0.7778$

• Part c: Suppose the odds against Texas A&M winning the national title are $8$ to $1$. Find the probability that Texas A&M will not win the national title.

  • Odds against winning are $8$ to $1$ $\implies$ Odds in favor of winning are $1$ to $8$ ($m = 1, n = 8$).

  • $P(\text{winning}) = \frac{1}{1 + 8} = \frac{1}{9}$

  • $P(\text{not winning}) = \frac{8}{1 + 8} = \frac{8}{9} \approx 0.8889$

• Part d: Find the odds in favor of rolling a sum of $8$ when two dice are rolled.

  • Outcomes resulting in a sum of $8$: $\{(2,6), (3,5), (4,4), (5,3), (6,2)\} \implies 5$ outcomes.

  • $P(\text{sum} = 8) = \frac{5}{36}$

  • $P(\text{sum} \neq 8) = 1 - \frac{5}{36} = \frac{31}{36}$

  • $\text{Odds in favor} = \frac{5/36}{31/36} = \frac{5}{31} \implies 5 \text{ to } 31 \text{ (or } 5:31)$

  • Example 5: Set Probabilities from Venn Diagrams

  Given $P(A) = 0.42$, $P(B) = 0.67$, and $P(A \cup B) = 0.81$. Find the following probabilities:

• Part a: $P(A \cap B)$

  • Rearranging the Union Rule:       $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.42 + 0.67 - 0.81 = 0.28$

• Part b: $P(B')$

  • $P(B') = 1 - P(B) = 1 - 0.67 = 0.33$

• Part c: $P(A \cap B')$

  • Event $A \cap B'$ represents outcomes in $A$ but not in $B$:       $P(A \cap B') = P(A) - P(A \cap B) = 0.42 - 0.28 = 0.14$

• Part d: $P(A' \cap B')$

  • Event $A' \cap B'$ is the complement of $A \cup B$ (by De Morgan's Law):       $P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.81 = 0.19$

  • Probability Keywords Translation:

• "and", "but" $\implies$ Intersection ($\cap$)

• "or", "either or" $\implies$ Union ($\cup$)

  • Example 6: Political Survey Probabilities   A survey of $150$ people was conducted regarding political views. $40$ people were older than $50$, $25$ were older than $50$ and Republican, and $105$ were Republican. Find the probability that a randomly chosen person:

• Let $O$ = Older than $50$ ($n(O) = 40$)

• Let $R$ = Republican ($n(R) = 105$)

• Let $O \cap R$ = Older than $50$ AND Republican ($n(O \cap R) = 25$)

• Part a: Is older than $50$ and a Republican ($P(O \cap R)$):     $P(O \cap R) = \frac{n(O \cap R)}{N} = \frac{25}{150} = \frac{1}{6} \approx 0.1667$

• Part b: Is older than $50$ or a Republican ($P(O \cup R)$):     $P(O \cup R) = P(O) + P(R) - P(O \cap R) = \frac{40}{150} + \frac{105}{150} - \frac{25}{150} = \frac{120}{150} = \frac{4}{5} = 0.80$

• Part c: Is not a Republican ($P(R')$):     $P(R') = 1 - P(R) = 1 - \frac{105}{150} = \frac{45}{150} = \frac{3}{10} = 0.30$

• Part d: Is older than $50$ but not a Republican ($P(O \cap R')$):     $P(O \cap R') = \frac{n(O) - n(O \cap R)}{N} = \frac{40 - 25}{150} = \frac{15}{150} = \frac{1}{10} = 0.10$

Contingency Tables and Empirical Probability

• Structure of Two-Way Contingency Tables:   A contingency table classifies sample observations according to two categorical variables (e.g., driver age group and accident frequency). The row events ($A, B, C, D, E$) are mutually exclusive and exhaustive. Similarly, the column events ($X, Y, Z$) are mutually exclusive and exhaustive.

• Driver Accident Data Summary Table:

  • Driver's Age Categories:

  • $A$: $16\text{--}19$ years old (Total = $50$)

  • $B$: $20\text{--}29$ years old (Total = $50$)

  • $C$: $30\text{--}39$ years old (Total = $50$)

  • $D$: $40\text{--}49$ years old (Total = $50$)

  • $E$: $50+$ years old (Total = $50$)

  • Accident Categories (Past Year):

  • $X$: $0$ accidents (Total = $181$)

  • $Y$: $1\text{--}2$ accidents (Total = $50$)

  • $Z$: $3+$ accidents (Total = $19$)

  • Total Sample Size ($N$): $250$

• Example 1: Probabilities from Contingency Table

  • Part a: Find the probability that a driver was $30\text{--}39$ years old ($C$) or had $3$ or more accidents ($Z$) in the past year.     $P(C \cup Z) = P(C) + P(Z) - P(C \cap Z)$     $P(C) = \frac{50}{250}, \quad P(Z) = \frac{19}{250}, \quad P(C \cap Z) = \frac{2}{250}$     $P(C \cup Z) = \frac{50 + 19 - 2}{250} = \frac{67}{250} = 0.268$

  • Part b: Find the probability that a driver was older than $19$ years old ($B \cup C \cup D \cup E$) or had not been in an accident in the past year ($X$).

  • Number of drivers older than $19$: $n(B \cup C \cup D \cup E) = 50 + 50 + 50 + 50 = 200$

  • Number of drivers with zero accidents: $n(X) = 181$

  • Overlap (drivers older than $19$ AND zero accidents): $n((B \cup C \cup D \cup E) \cap X) = 35 + 40 + 42 + 39 = 156$

  • Calculation:       $P = \frac{200 + 181 - 156}{250} = \frac{225}{250} = \frac{9}{10} = 0.90$

  • Part c: Find the probability that a driver was $20\text{--}29$ years old ($B$) or older than $50$ ($E$).

  • Since age categories $B$ and $E$ are mutually exclusive ($B \cap E = \emptyset$):       $P(B \cup E) = P(B) + P(E) = \frac{50}{250} + \frac{50}{250} = \frac{100}{250} = \frac{2}{5} = 0.40$

Conditional Probability and Probability Trees

• Definition of Conditional Probability:   The conditional probability of event $E$ occurring, given that event $F$ has already occurred, is denoted $P(E|F)$ and defined as:   $P(E|F) = \frac{P(E \cap F)}{P(F)}, \quad P(F) \neq 0$   Alternatively, using event counts from a finite sample space:   $P(E|F) = \frac{n(E \cap F)}{n(F)}, \quad n(F) \neq 0$

• Example 1: Conditional Probability from Venn Diagram Data   Consider a universal sample space divided into four regions $a, b, c, d$ with given probabilities:

  • Region $a$ ($A \cap B'$) = $0.20$

  • Region $b$ ($A \cap B$) = $0.30$

  • Region $c$ ($A' \cap B$) = $0.35$

  • Region $d$ ($A' \cap B'$) = $0.15$

  • Marginal probabilities: $P(A) = 0.20 + 0.30 = 0.50$, $P(B) = 0.30 + 0.35 = 0.65$

  • Part a: $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.30}{0.65} = \frac{6}{13} \approx 0.4615$

  • Part b: $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.30}{0.50} = \frac{3}{5} = 0.60$

  • Part c: $P(A'|B) = \frac{P(A' \cap B)}{P(B)} = \frac{0.35}{0.65} = \frac{7}{13} \approx 0.5385$

  • Part d: $P(B'|A') = \frac{P(A' \cap B')}{P(A')} = \frac{0.15}{1 - 0.50} = \frac{0.15}{0.50} = \frac{3}{10} = 0.30$

• Product Rule of Probability:   By rearranging the conditional probability formula, the probability of the intersection of two events $E$ and $F$ is given by:   $P(E \cap F) = P(F) \cdot P(E|F) \quad \text{or} \quad P(E \cap F) = P(E) \cdot P(F|E)$

• Example 2: Tree Diagram with Marble Selection (Without Replacement)   A box contains $3$ red ($R$), $6$ black ($B$), and $1$ white ($W$) marble (Total = $10$ marbles). Two marbles are drawn sequentially without replacement.

  • Part a: Find the probability that the first marble is black ($B_1$).     $P(B_1) = \frac{6}{10} = \frac{3}{5} = 0.60$

  • Part b: Find the probability that the second marble is black given that the first marble was white ($P(B_2 | W_1)$).

  • If the first marble drawn was white, $9$ marbles remain in the box, $6$ of which are black:       $P(B_2 | W_1) = \frac{6}{9} = \frac{2}{3} \approx 0.6667$

  • Part c: Find the probability that both marbles are red ($P(R_1 \cap R_2)$).     $P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 | R_1) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15} \approx 0.0667$

  • Part d: Find the probability that one marble is red and one marble is white (in any order).

  • Two mutually exclusive paths exist: $(R_1 \cap W_2)$ or $(W_1 \cap R_2)$.       $P(R_1 \cap W_2) = P(R_1) \cdot P(W_2 | R_1) = \frac{3}{10} \times \frac{1}{9} = \frac{3}{90}$       $P(W_1 \cap R_2) = P(W_1) \cdot P(R_2 | W_1) = \frac{1}{10} \times \frac{3}{9} = \frac{3}{90}$       $P(\text{one } R \text{ and one } W) = \frac{3}{90} + \frac{3}{90} = \frac{6}{90} = \frac{1}{15} \approx 0.0667$

  • Part e: Find the probability that the second marble is red ($P(R_2)$).

  • Sum the probabilities of all pathways leading to a red marble on the second draw ($R_1 R_2$, $B_1 R_2$, $W_1 R_2$):       $P(R_2) = P(R_1 \cap R_2) + P(B_1 \cap R_2) + P(W_1 \cap R_2)$       $P(R_2) = \left(\frac{3}{10} \times \frac{2}{9}\right) + \left(\frac{6}{10} \times \frac{3}{9}\right) + \left(\frac{1}{10} \times \frac{3}{9}\right) = \frac{6 + 18 + 3}{90} = \frac{27}{90} = \frac{3}{10} = 0.30$

• Example 3: Sequential Card Selection   Two cards are drawn sequentially without replacement from a standard deck of $52$ cards.

  • Part a: Probability that the second card is a club given that the first card is a club.

  • Remaining clubs = $12$, remaining cards = $51$:       $P(\text{Club}_2 | \text{Club}_1) = \frac{12}{51} = \frac{4}{17} \approx 0.2353$

  • Part b: Probability that both cards are diamonds.     $P(D_1 \cap D_2) = P(D_1) \cdot P(D_2 | D_1) = \frac{13}{52} \times \frac{12}{51} = \frac{1}{4} \times \frac{12}{51} = \frac{3}{51} = \frac{1}{17} \approx 0.0588$

  • Part c: Probability that a spade and a heart are drawn in any order.     $P((S_1 \cap H_2) \cup (H_1 \cap S_2)) = \left(\frac{13}{52} \times \frac{13}{51}\right) + \left(\frac{13}{52} \times \frac{13}{51}\right) = 2 \times \left(\frac{1}{4} \times \frac{13}{51}\right) = \frac{13}{102} \approx 0.1275$

• Example 4: Voter Preference Tree Application   In a community, $40\%$ of voters are Republican ($R$), $25\%$ are Democrats ($D$), and $35\%$ are Independents ($I$). Voting preference for Candidate A ($A$):

  • $P(A|R) = 0.90$

  • $P(A|D) = 0.15$

  • $P(A|I) = 0.75$

  • Part a: Probability that a registered voter was a Republican AND voted for Candidate A:     $P(R \cap A) = P(R) \cdot P(A|R) = 0.40 \times 0.90 = 0.36 = 36\%$

  • Part b: Total percent of registered voters who voted for Candidate A ($P(A)$):     $P(A) = P(R \cap A) + P(D \cap A) + P(I \cap A)$     $P(A) = (0.40)(0.90) + (0.25)(0.15) + (0.35)(0.75) = 0.36 + 0.0375 + 0.2625 = 0.66 = 66\%$

  • Part c: Percent of registered voters who were Republican OR voted for Candidate A ($P(R \cup A)$):     $P(R \cup A) = P(R) + P(A) - P(R \cap A) = 0.40 + 0.66 - 0.36 = 0.70 = 70\%$

• Example 5: Conditional Probabilities from Contingency Table   Using the driver accident table ($N = 250$):

  • Part a: Driver was $30\text{--}39$ years old ($C$) given that the driver did NOT have an accident in the past year ($X$):     $P(C|X) = \frac{n(C \cap X)}{n(X)} = \frac{40}{181} \approx 0.2210$

  • Part b: Driver had $1\text{--}2$ accidents ($Y$) knowing that the driver was $50$ or older ($E$):     $P(Y|E) = \frac{n(Y \cap E)}{n(E)} = \frac{10}{50} = \frac{1}{5} = 0.20$

  • Part c: Driver was $20\text{--}29$ years old ($B$) if he or she had been in at least one accident ($Y \cup Z$):     $n(Y \cup Z) = 50 + 19 = 69, \quad n(B \cap (Y \cup Z)) = 12 + 3 = 15$     $P(B | Y \cup Z) = \frac{15}{69} = \frac{5}{23} \approx 0.2174$

  • Part d: Probability a teenage driver ($A$) had $3$ or more accidents ($Z$) [Conditional $P(Z|A)$]:     $P(Z|A) = \frac{n(A \cap Z)}{n(A)} = \frac{10}{50} = \frac{1}{5} = 0.20$

  • Part e: Driver was a teenager AND had $3$ or more accidents [Joint $P(A \cap Z)$]:     $P(A \cap Z) = \frac{n(A \cap Z)}{N} = \frac{10}{250} = \frac{1}{25} = 0.04$

Independence of Events

• Definition of Independent Events:   Two events $E$ and $F$ are independent if the occurrence of one event has no effect on the probability of the occurrence of the other. Formally, $E$ and $F$ are independent if and only if:   $P(E|F) = P(E) \quad \text{or} \quad P(F|E) = P(F)$   If events are not independent, they are defined as dependent events.

• Product Rule for Independent Events:   Events $E$ and $F$ are independent if and only if:   $P(E \cap F) = P(E) \cdot P(F)$

• Example 1: Conceptual Identification of Independence

  • Scenario a: Draw two cards from a deck of $52$ without replacement. First card is a diamond, second card is red.

  • Dependent: Drawing the first card without replacement alters the total number of cards and red cards remaining, affecting the probability of the second draw.

  • Scenario b: The Houston Astros lose a baseball game and the Dallas Cowboys win a football game.

  • Independent: The outcome of one sports team's game has no direct mathematical effect on the outcome of an unrelated game by another team.

  • Scenario c: Roll two dice. First die shows a $5$, second die shows a $3$.

  • Independent: The outcome of the first die roll does not physically or probabilistically influence the outcome of the second die roll.

• Example 2: Mathematical Test for Independence   Given events $A$ and $B$ such that $P(A) = 0.30$, $P(B) = 0.40$, and $P(A \cap B) = 0.12$. Determine if $A$ and $B$ are independent.

  • Check product condition: $P(A) \cdot P(B) = (0.30)(0.40) = 0.12$

  • Since $P(A \cap B) = P(A) \cdot P(B)$, events $A$ and $B$ are independent.

• Example 3: Weather and Traffic Analysis   On a typical January day, $P(\text{snow}) = P(S) = 0.10$, $P(\text{traffic jam}) = P(T) = 0.80$, and $P(S \cup T) = 0.82$. Are snowing and traffic jams independent?

  • First, calculate $P(S \cap T)$ using the Union Rule:     $P(S \cup T) = P(S) + P(T) - P(S \cap T)$     $0.82 = 0.10 + 0.80 - P(S \cap T) \implies 0.82 = 0.90 - P(S \cap T) \implies P(S \cap T) = 0.08$

  • Test product rule:     $P(S) \cdot P(T) = (0.10)(0.80) = 0.08$

  • Since $P(S \cap T) = P(S) \cdot P(T) = 0.08$, the events are independent.

• Example 4: Dice Roll Events Independence Test   Roll two fair dice. Let $E$ = sum of eight is rolled, and $F$ = exactly one $2$ is rolled. Are $E$ and $F$ independent?

  • $E = \{(2,6), (3,5), (4,4), (5,3), (6,2)\} \implies n(E) = 5 \implies P(E) = \frac{5}{36}$

  • $F = \{(2,1), (2,3), (2,4), (2,5), (2,6), (1,2), (3,2), (4,2), (5,2), (6,2)\} \implies n(F) = 10 \implies P(F) = \frac{10}{36} = \frac{5}{18}$

  • Intersection $E \cap F = \{(2,6), (6,2)\} \implies n(E \cap F) = 2 \implies P(E \cap F) = \frac{2}{36} = \frac{1}{18}$

  • Test product rule:     $P(E) \cdot P(F) = \frac{5}{36} \times \frac{10}{36} = \frac{50}{1296} = \frac{25}{648} \approx 0.0386$

  • Compare $P(E \cap F) = \frac{2}{36} = \frac{36}{648} \approx 0.0556$

  • Since $P(E \cap F) \neq P(E) \cdot P(F)$, the events are dependent.

• Example 5: Independence Test from Contingency Table   From the driver accident table, test if event $E$ (driver is $50$ or older) and event $Y$ (driver had $1\text{--}2$ accidents) are independent.

  • $P(E) = \frac{50}{250} = 0.20$

  • $P(Y) = \frac{50}{250} = 0.20$

  • $P(E \cap Y) = \frac{10}{250} = 0.04$

  • Test product rule:     $P(E) \cdot P(Y) = (0.20)(0.20) = 0.04$

  • Since $P(E \cap Y) = P(E) \cdot P(Y)$, the events are independent.

Random Variables and Probability Distributions

• Definition of Random Variable:   A random variable $X$ is a function that assigns a unique real number to each outcome of a probability experiment. Random variables are categorized into three types:

  • Finite Discrete: The random variable can take on a finite number of distinct values (e.g., counting the number of aces in a $5$-card hand).

  • Infinite Discrete: The random variable can take on an infinite, countable sequence of distinct values (e.g., rolling a die repeatedly until a six appears and counting the number of rolls required).

  • Continuous: The random variable can take on any real number within a continuous interval (e.g., measuring the distance between two arbitrary points on Earth).

• Example 1: Random Variables and Outcome Sets

  • Scenario a: Draw $2$ marbles from a bag containing $10$ red, $6$ blue, and $2$ white marbles without replacement. Let $X$ be the number of red marbles drawn.

  • Possible values for $X$: $X \in \{0, 1, 2\}$

  • Scenario b1: Roll two fair six-sided dice. Let $X$ be the number of sixes rolled.

  • Possible values for $X$: $X \in \{0, 1, 2\}$

  • Scenario b2: Roll two fair six-sided dice. Let $X$ be the sum of the dice.

  • Possible values for $X$: $X \in \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$

• Definition of Probability Distribution:   A probability distribution is a table, graph, or formula that lists all possible numerical values $x_i$ of a discrete random variable $X$ alongside their corresponding probabilities $P(X = x_i)$, satisfying:   $\sum P(X = x_i) = 1 \quad \text{and} \quad 0 \le P(X = x_i) \le 1$

• Example 2: Constructing Probability Distribution Tables

  • Part a: Drawing $2$ marbles from $10$ red, $6$ blue, and $2$ white (Total = $18$ marbles). Let $X$ = number of red marbles.

  • Total ways to choose $2$ marbles from $18$: $\binom{18}{2} = \frac{18 \times 17}{2} = 153$

  • $X = 0$ (Choose $2$ non-red from $8$): $\binom{8}{2} = 28 \implies P(X = 0) = \frac{28}{153}$

  • $X = 1$ (Choose $1$ red from $10$, $1$ non-red from $8$): $\binom{10}{1} \binom{8}{1} = 10 \times 8 = 80 \implies P(X = 1) = \frac{80}{153}$

  • $X = 2$ (Choose $2$ red from $10$): $\binom{10}{2} = 45 \implies P(X = 2) = \frac{45}{153}$

  • Distribution Table:       \begin{array}{|c|c|c|c|}\n      \hline\n      X & 0 & 1 & 2 \\\n      \hline\n      P(X) & \frac{28}{153} & \frac{80}{153} & \frac{45}{153} \\\n      \hline\n      \end{array}

  • Part b1: Roll two fair dice. Let $X$ = number of sixes rolled.

  • $X = 0$ (Neither die is $6$): $\frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$

  • $X = 1$ (One die is $6$): $2 \times \left(\frac{1}{6} \times \frac{5}{6}\right) = \frac{10}{36}$

  • $X = 2$ (Both dice are $6$): $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

  • Distribution Table:       \begin{array}{|c|c|c|c|}\n      \hline\n      X & 0 & 1 & 2 \\\n      \hline\n      P(X) & \frac{25}{36} & \frac{10}{36} & \frac{1}{36} \\\n      \hline\n      \end{array}

  • Part b2: Roll two fair dice. Let $X$ = sum of the dice.

  • Distribution Table:       \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}\n      \hline\n      X & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\\n      \hline\n      P(X) & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{6}{36} & \frac{5}{36} & \frac{4}{36} & \frac{3}{36} & \frac{2}{36} & \frac{1}{36} \\\n      \hline\n      \end{array}

Expected Value and Decision Analysis

• Definition and Formula for Expected Value:   Suppose a discrete random variable $X$ can take on values $x_1, x_2, \dots, x_n$ with corresponding probabilities $p_1, p_2, \dots, p_n$. The expected value $E(X)$ (or mean $\mu$) is the weighted average outcome over long-term repetition:   $E(X) = x_1 p_1 + x_2 p_2 + \dots + x_n p_n = \sum_{i=1}^{n} x_i P(X = x_i)$

• Example 1: Expected Value Computations

  • Part a: Find $E(X)$ for the marble selection experiment:     $E(X) = (0)\left(\frac{28}{153}\right) + (1)\left(\frac{80}{153}\right) + (2)\left(\frac{45}{153}\right) = \frac{0 + 80 + 90}{153} = \frac{170}{153} = \frac{10}{9} \approx 1.111$

  • Part b: Find $E(X)$ for the sum of two fair dice:     $E(X) = (2)\left(\frac{1}{36}\right) + (3)\left(\frac{2}{36}\right) + (4)\left(\frac{3}{36}\right) + (5)\left(\frac{4}{36}\right) + (6)\left(\frac{5}{36}\right) + (7)\left(\frac{6}{36}\right) + (8)\left(\frac{5}{36}\right) + (9)\left(\frac{4}{36}\right) + (10)\left(\frac{3}{36}\right) + (11)\left(\frac{2}{36}\right) + (12)\left(\frac{1}{36}\right)$     $E(X) = \frac{2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12}{36} = \frac{252}{36} = 7$

• Games of Chance and Decision Rules:   Expected value determines long-term profitability in decision analysis:

  • If $E(X) > 0$ (Positive expected winnings): The player should play.

  • If $E(X) < 0$ (Negative expected winnings): The player should not play.

  • Fair Game: A game is mathematically fair if and only if its expected value is zero ($E(X) = 0$).

• Example 2: Card Game Expectation Between Two Players   Player A draws a card from a deck of $52$. If a face card (Jack, Queen, King; $12$ total) is drawn, Player A wins \$80$. If any other card ($40$ total) is drawn, Player A loses \20$.\n * Random Variable X = Player A's net payoff.\n * Distribution Table:\n    \begin{array}{|c|c|c|}     \hline     X & +80 & -20 \     \hline     P(X) & \frac{12}{52} = \frac{3}{13} & \frac{40}{52} = \frac{10}{13} \     \hline     \end{array}\n * Expected value per game:\n    E(X) = (80)\left(\frac{3}{13}\right) + (-20)\left(\frac{10}{13}\right) = \frac{240 - 200}{13} = \frac{40}{13} \approx \$3.08\n * Expected status after 10 games:\n    \text{Expected Winnings} = 10 \times \frac{40}{13} = \frac{400}{13} \approx \30.77 \text{ (Up } \30.77)\n\n* **Example 3: Raffle Prize Expected Winnings**\n  Frank pays \1$to draw a ball from a barrel containing$4000 total balls. Prizes contained in the balls:\n * One \500 gift card\n * Four \100 gift cards\n * Ten \50 gift cards\n * Remaining 3985 balls contain no prize (\0).\n * Compute net winnings X (Prize minus \1 entry fee):\n * Net \499$:$P = \frac{1}{4000}\n * Net \99$:$P = \frac{4}{4000}\n * Net \49$:$P = \frac{10}{4000}\n * Net \-1$:$P = \frac{3985}{4000}\n * Expected value equation:\n    E(X) = (499)\left(\frac{1}{4000}\right) + (99)\left(\frac{4}{4000}\right) + (49)\left(\frac{10}{4000}\right) + (-1)\left(\frac{3985}{4000}\right)\n    E(X) = \frac{499 + 396 + 490 - 3985}{4000} = \frac{-2600}{4000} = -\$0.65\n * Frank loses an average of \0.65 per drawing.\n\n* **Example 4: Casino Roulette Analysis**\n\n\n\n  An American roulette wheel contains 38$pockets ($18$Red,$18$Black,$2 Green). Sally places a \100 bet on Red and a \50 bet on Green (Total outlay = \150).\n * **Payout Rules**:\n * Red landing (18/38): Wins back \100 bet + \100 profit on Red. Loses \50$Green bet. Net profit =$+100 - 50 = +\$50.\n * Green landing (2/38): Wins back \50$bet +$20 \times 50 = \$1000 profit on Green. Loses \100$Red bet. Net profit =$+1000 - 100 = +\$900.\n * Black landing (18/38$): Loses both bets. Net profit =$-\$150.\n * Expected Winnings E(X):\n    E(X) = (50)\left(\frac{18}{38}\right) + (900)\left(\frac{2}{38}\right) + (-150)\left(\frac{18}{38}\right)\n    E(X) = \frac{900 + 1800 - 2700}{38} = \frac{0}{38} = \$0\n * **Fairness**: The game is **fair** because E(X) = \$0.\n\n* **Example 5: Spinner Game Analysis**\n\n\n\n  Hank spins an 8$-sector spinner twice. The spinner has$5$happy-face sectors ($S$) and$3$frowny-face sectors ($F).\n * Probabilities for a single spin: P(S) = \frac{5}{8}$,$P(F) = \frac{3}{8}\n * Two-spin probabilities:\n * Two happy faces (SS$):$P(SS) = \frac{5}{8} \times \frac{5}{8} = \frac{25}{64}, Net win = \50\n * Exactly one happy face (SF \text{ or } FS$):$P(1S) = 2 \times \left(\frac{5}{8} \times \frac{3}{8}\right) = \frac{30}{64}, Net win = \10\n * Two frowny faces (FF$):$P(FF) = \frac{3}{8} \times \frac{3}{8} = \frac{9}{64}, Net loss = \-100\n * **Expected Value for one game**:\n    E(X) = (50)\left(\frac{25}{64}\right) + (10)\left(\frac{30}{64}\right) + (-100)\left(\frac{9}{64}\right)\n    E(X) = \frac{1250 + 300 - 900}{64} = \frac{650}{64} = \frac{325}{32} \approx \$10.16\n * **Expected status after 20 games**:\n    \text{Total Expected Winnings} = 20 \times \frac{650}{64} = \frac{13000}{64} = \frac{1625}{8} = \203.125 \approx \203.13\n * **Fairness**: The game is **not fair** because E(X) eq 0$($E(X) \approx \$10.16$$).