Developing Theoretical Probability Distributions, Expected Value - Statistics and Probability

Two fair coins are flipped. The equally likely outcomes are $\{HH,HT,TH,TT\}$ . Define the random variable $X$ as the number of heads obtained. Which table correctly represents the theoretical probability distribution of $X$ ?
Answer: $P(X=0)=\tfrac14,\;P(X=1)=\tfrac12,\;P(X=2)=\tfrac14$ This problem requires developing a theoretical probability distribution for counting heads in two coin flips. The four equally likely outcomes {HH,HT,TH,TT} each have probability 1/4. We map these to X-values by counting heads: HH → X=2, HT → X=1, TH → X=1, TT → X=0. To find P(X=x), we combine probabilities of outcomes with the same head count: P(X=0) = P(TT) = 1/4, P(X=1) = P(HT) + P(TH) = 1/4 + 1/4 = 1/2, P(X=2) = P(HH) = 1/4. The distribution is not uniform across X-values because multiple outcomes map to X=1. The transfer strategy is: list all outcomes → count heads for each → combine probabilities for same counts.
A fair coin is flipped 3 times. All $8$ outcomes in the sample space are equally likely. Define the random variable $X$ as the number of heads minus the number of tails. Explicit mapping from outcomes to $X$ : - $\text{HHH}\mapsto 3$ - $\text{HHT, HTH, THH}\mapsto 1$ - $\text{HTT, THT, TTH}\mapsto -1$ - $\text{TTT}\mapsto -3$ Which table correctly represents the theoretical probability distribution of $X$ (probabilities derived from the sample space, not data)?
Answer: $P(X=3)=\frac{1}{8},\ P(X=1)=\frac{3}{8},\ P(X=-1)=\frac{3}{8},\ P(X=-3)=\frac{1}{8}$ This problem involves developing theoretical probability distributions and expected value, here focusing on the probability distribution. Probabilities are derived from the uniform model over the sample space of 8 equally likely outcomes for three fair coin flips. The outcomes are mapped to X-values as the number of heads minus the number of tails, resulting in X=3 for HHH, X=1 for two heads, X=-1 for one head, and X=-3 for TTT. The probability P(X=x) is determined by counting the outcomes mapping to each x and dividing by 8, yielding P(X=3)=1/8, P(X=1)=3/8, P(X=-1)=3/8, and P(X=-3)=1/8. This distribution can be used to compute expected value if needed, but here we identify the correct table. A common misconception is assuming all X-values are equally likely, but values like 1 and -1 are three times more probable than 3 or -3. To solve similar problems, list all outcomes, map each to X, and combine probabilities for each distinct X-value to build the distribution.
A game uses a bag containing 5 equally likely tickets labeled $A,B,C,D,E$ . One ticket is drawn at random. Define the random variable $X$ by the mapping: $X=-2$ if $A$ is drawn; $X=0$ if $B$ or $C$ is drawn; and $X=3$ if $D$ or $E$ is drawn. What is the expected value of $X$ based on this theoretical model?
Answer: $\dfrac{4}{5}$ This problem involves developing a theoretical probability distribution from a bag of tickets. The model has 5 equally likely tickets {A,B,C,D,E}, each with probability 1/5. The random variable X maps: X=-2 when A is drawn, X=0 when B or C is drawn, and X=3 when D or E is drawn. To find P(X=x), we count tickets: P(X=-2) = P(A) = 1/5, P(X=0) = P(B)+P(C) = 2/5, and P(X=3) = P(D)+P(E) = 2/5. The expected value is E(X) = (-2)(1/5) + 0(2/5) + 3(2/5) = -2/5 + 0 + 6/5 = 4/5. A common misconception is thinking X-values {-2,0,3} are equally likely with probability 1/3 each, or averaging the X-values directly as (3+0+(-2))/3. The strategy: identify which tickets give each X-value → count them → calculate probabilities as fractions of 5.
Two fair coins are flipped once each, producing 4 equally likely outcomes: $\{\text{HH, HT, TH, TT}\}$ . Define the random variable $X$ by the explicit mapping: $\text{HH}\mapsto 2$ , $\text{HT}\mapsto 0$ , $\text{TH}\mapsto 0$ , $\text{TT}\mapsto -1$ . Which table correctly represents the theoretical probability distribution of $X$ ?
Answer: $P(X=-1)=\tfrac{1}{4},\ P(X=0)=\tfrac{1}{2},\ P(X=2)=\tfrac{1}{4}$ This problem involves developing theoretical probability distributions and calculating expected values for discrete random variables. In this model, probabilities are derived from the four equally likely outcomes of two fair coin flips, each with probability 1/4. Each outcome maps explicitly to an X-value: HH to 2, HT to 0, TH to 0, and TT to -1. The probability P(X=x) is determined by grouping outcomes: P(X=2)=1/4, P(X=0)=2/4=1/2, P(X=-1)=1/4. The expected value E(X) is computed as 2*(1/4) + 0*(1/2) + (-1)*(1/4) = (2/4 + 0 - 1/4) = 1/4. A common misconception is that all X-values are equally likely, but P(X=0) is twice as likely due to more mappings. To solve similar problems, list all outcomes, map each to its X, and combine probabilities for each distinct x.
A fair six-sided die is rolled once. All outcomes are equally likely. Define the random variable $X$ by mapping outcomes to values as follows: $\{1,2,3\}\mapsto 0$ , $\{4,5\}\mapsto 1$ , and $\{6\}\mapsto 3$ . Which table correctly represents the theoretical probability distribution of $X$ ?
Answer: $P(X=0)=\dfrac{3}{6},\ P(X=1)=\dfrac{2}{6},\ P(X=3)=\dfrac{1}{6}$ This problem involves finding the theoretical probability distribution of X when a fair six-sided die maps outcomes to X-values. The die outcomes {1,2,3,4,5,6} each have probability 1/6. The random variable X maps outcomes as follows: outcomes {1,2,3} all map to X = 0, outcomes {4,5} both map to X = 1, and outcome {6} maps to X = 3. To find P(X = 0), we combine probabilities of outcomes mapping to 0: P(X = 0) = P(1) + P(2) + P(3) = 1/6 + 1/6 + 1/6 = 3/6. Similarly, P(X = 1) = P(4) + P(5) = 1/6 + 1/6 = 2/6, and P(X = 3) = P(6) = 1/6. The distribution is P(X = 0) = 3/6, P(X = 1) = 2/6, P(X = 3) = 1/6. A common misconception is thinking X-values 0, 1, and 3 are equally likely, each with probability 1/3, without considering how many die outcomes map to each X-value.
A fair six-sided die is rolled once (outcomes $1$ through $6$ are equally likely). Define the random variable $X$ by the mapping: $1\mapsto -2$ , $2\mapsto -2$ , $3\mapsto 0$ , $4\mapsto 0$ , $5\mapsto 3$ , $6\mapsto 9$ . What is the expected value of $X$ based on this theoretical model?
Answer: $E(X)=\dfrac{4}{3}$ This problem maps die outcomes to specific values to find expected value. The six equally likely outcomes each have probability 1/6. The mapping creates a distribution where: X=-2 with probability 2/6 (outcomes 1,2), X=0 with probability 2/6 (outcomes 3,4), X=3 with probability 1/6 (outcome 5), and X=9 with probability 1/6 (outcome 6). The expected value is E(X) = (-2)(2/6) + 0(2/6) + 3(1/6) + 9(1/6) = -4/6 + 0 + 3/6 + 9/6 = 8/6 = 4/3. A common error is averaging the four distinct X-values (-2,0,3,9)/4 without considering that -2 and 0 each occur twice. The key: track how many die outcomes produce each X-value.
A fair six-sided die is rolled once. The sample space is $\{1,2,3,4,5,6\}$ with all outcomes equally likely. Define the random variable $X$ as the number of points you win where you win 2 points if the roll is even and 0 points if the roll is odd. (So outcomes $\{2,4,6\}$ map to $X=2$ and outcomes $\{1,3,5\}$ map to $X=0$ .) What is the expected value of $X$ based on this theoretical model?
Answer: $1$ This problem focuses on theoretical probability distributions and expected value in the context of a fair die roll. Probabilities are derived from the uniform model where each of the six outcomes is equally likely with probability 1/6. Outcomes map to X-values such that even numbers {2,4,6} correspond to X=2 and odd numbers {1,3,5} correspond to X=0. The probabilities are determined by grouping: P(X=2) = 3/6 = 1/2 and P(X=0) = 3/6 = 1/2. The expected value is computed as E[X] = 2*(1/2) + 0*(1/2) = 1. A common misconception is that distinct X-values are always equally likely, but here they are due to balanced groupings, whereas uneven groupings would yield different probabilities. To approach similar problems, list all equally likely outcomes, map them to X-values, and combine probabilities for each unique X.
A game uses a fair spinner with 6 equally likely outcomes labeled $1,2,3,4,5,6$ . Define the random variable $X$ as the payoff (in dollars) where outcomes $1$ or $2$ pay $0$ , outcomes $3$ or $4$ pay $2$ , outcome $5$ pays $5$ , and outcome $6$ pays $8$ . What is the expected value of $X$ based on this theoretical model?
Answer: $E(X)=\dfrac{17}{6}$ This problem involves theoretical probability distributions and expected value calculations. The spinner has 6 equally likely outcomes, each with probability 1/6. We map these outcomes to payoff values: outcomes 1,2 → X=0; outcomes 3,4 → X=2; outcome 5 → X=5; outcome 6 → X=8. To find P(X=x), we combine probabilities of outcomes mapping to the same X-value: P(X=0) = 2/6, P(X=2) = 2/6, P(X=5) = 1/6, P(X=8) = 1/6. The expected value is E(X) = 0(2/6) + 2(2/6) + 5(1/6) + 8(1/6) = 0 + 4/6 + 5/6 + 8/6 = 17/6. A common misconception is assuming X-values 0,2,5,8 are equally likely, but we must consider how many spinner outcomes map to each payoff.
A fair six-sided die is rolled once; outcomes are equally likely. Define the random variable $X$ as $X=\lvert\text{roll}-3\rvert$ . The outcome-to- $X$ mapping is: $1\to 2$ , $2\to 1$ , $3\to 0$ , $4\to 1$ , $5\to 2$ , $6\to 3$ . Which table correctly represents the theoretical probability distribution of $X$ ?
Answer: $P(X=0)=\frac{1}{6},\;P(X=1)=\frac{2}{6},\;P(X=2)=\frac{2}{6},\;P(X=3)=\frac{1}{6}$ This problem covers theoretical probability distributions and expected value for absolute deviation on a die. Probabilities come from the uniform model with six equally likely outcomes. Outcomes map to X = |roll - 3|, yielding 3 to 0, 2 and 4 to 1, 1 and 5 to 2, and 6 to 3. Hence, P(X=0) = 1/6, P(X=1) = 2/6, P(X=2) = 2/6, P(X=3) = 1/6, as in choice A. This supports E[X] = 0*(1/6) + 1*(2/6) + 2*(2/6) + 3*(1/6) = 4/3. A misconception is equal probability for each X, but it varies with mapped outcomes. For like problems, list outcomes, compute X for each, and aggregate probabilities per X-value.
A fair spinner is divided into 5 equal sectors labeled $1,2,3,4,5$ , so each outcome is equally likely. Define the random variable $X$ by the mapping: $\{1,2,3\}\mapsto -2$ and $\{4,5\}\mapsto 4$ . What is the expected value of $X$ based on this theoretical model?
Answer: $E(X)=\dfrac{2}{5}$ This problem involves developing theoretical probability distributions and calculating expected value. Probabilities are derived from the uniform model over five equally likely sectors on a fair spinner labeled 1 through 5. The outcomes are mapped to X-values: {1,2,3} map to -2, and {4,5} map to 4. The probability P(X=-2) is the sum of probabilities for 1,2,3, which is 3/5, and P(X=4) = 2/5. The expected value is computed as E(X) = (-2) (3/5) + 4 (2/5) = 2/5. A common misconception is assuming X-values are equally likely, but here P(-2) is larger due to more outcomes mapping to it. To solve similar problems, list all outcomes, map each to X, and combine probabilities for each distinct X to find the distribution and expected value.
A fair six-sided die is rolled once (outcomes $1$ – $6$ equally likely). Define the random variable $X$ by the explicit mapping: $X=1$ if the outcome is $1$ or $2$ ; $X=2$ if the outcome is $3$ ; $X=3$ if the outcome is $4$ or $5$ ; and $X=6$ if the outcome is $6$ . Which table correctly represents the theoretical probability distribution of $X$ ?
Answer: $P(X=1)=\dfrac{2}{6},\;P(X=2)=\dfrac{1}{6},\;P(X=3)=\dfrac{2}{6},\;P(X=6)=\dfrac{1}{6}$ This problem requires developing a theoretical probability distribution from an explicit die mapping. The model has 6 equally likely outcomes {1,2,3,4,5,6}, each with probability 1/6. The random variable X maps: X=1 when outcome is 1 or 2, X=2 when outcome is 3, X=3 when outcome is 4 or 5, and X=6 when outcome is 6. To find P(X=x), we count die outcomes: P(X=1) = P(1)+P(2) = 2/6, P(X=2) = P(3) = 1/6, P(X=3) = P(4)+P(5) = 2/6, and P(X=6) = P(6) = 1/6. The expected value would be E(X) = 1(2/6) + 2(1/6) + 3(2/6) + 6(1/6) = 2/6 + 2/6 + 6/6 + 6/6 = 16/6 = 8/3. A common misconception is thinking the four X-values {1,2,3,6} are equally likely, ignoring that some have multiple die outcomes mapping to them. The strategy: identify die outcomes for each X-value → count them → express probabilities as fractions of 6.
A fair six-sided die is rolled once (each of $1$ – $6$ equally likely). Define the random variable $X$ by the mapping: odd outcomes $\{1,3,5\}\mapsto 1$ and even outcomes $\{2,4,6\}\mapsto 4$ . Which table correctly represents the theoretical probability distribution of $X$ ?
Answer: $P(X=1)=\tfrac{1}{2},\ P(X=4)=\tfrac{1}{2}$ This problem involves developing theoretical probability distributions and calculating expected values for discrete random variables. In this model, probabilities are derived from the six equally likely outcomes of a fair die roll, each with probability 1/6. Outcomes map to X-values: odd numbers (1,3,5) to 1 and even numbers (2,4,6) to 4. The probability P(X=x) is determined by counting outcomes per x, giving P(X=1)=3/6=1/2 and P(X=4)=3/6=1/2. The expected value E(X) is computed as 1*(1/2) + 4*(1/2) = (1/2 + 4/2) = 5/2. A common misconception is that all X-values are equally likely only if outcome counts match, but here they are equal due to balanced mappings. To solve similar problems, list all outcomes, map each to its X, and combine probabilities for each distinct x.
Three fair coins are flipped. The $8$ outcomes are equally likely. Define the random variable $X$ as the number of tails obtained. Which table correctly represents the theoretical probability distribution of $X$ ?
Answer: $P(X=0)=\tfrac18,\;P(X=1)=\tfrac38,\;P(X=2)=\tfrac38,\;P(X=3)=\tfrac18$ This problem develops the probability distribution for counting tails in three coin flips. The 8 equally likely outcomes each have probability 1/8. Counting tails: TTT→3, TTH→2, THT→2, HTT→2, THH→1, HTH→1, HHT→1, HHH→0. Grouping by tail count: P(X=0) = 1/8 (one outcome: HHH), P(X=1) = 3/8 (three outcomes: THH, HTH, HHT), P(X=2) = 3/8 (three outcomes: TTH, THT, HTT), P(X=3) = 1/8 (one outcome: TTT). This follows the binomial pattern with symmetry around X=1.5. A misconception is thinking all values 0,1,2,3 are equally likely, but the middle values occur more frequently. The pattern matches Pascal's triangle row 3: (1,3,3,1).
A fair four-sided die with faces $\{1,2,3,4\}$ is rolled twice. All 16 ordered outcomes $(a,b)$ are equally likely. Define the random variable $X=\max(a,b)$ (the larger of the two rolls). Which table correctly represents the theoretical probability distribution of $X$ ?
Answer: $P(X=1)=\dfrac{1}{16},\ P(X=2)=\dfrac{3}{16},\ P(X=3)=\dfrac{5}{16},\ P(X=4)=\dfrac{7}{16}$ This problem involves finding the theoretical probability distribution of X = max(a,b) when rolling a fair four-sided die twice. The sample space contains 16 equally likely ordered pairs (a,b), each with probability 1/16. To find P(X = k), we count pairs where the maximum equals k. For X = 1, both rolls must be 1: only (1,1), so P(X = 1) = 1/16. For X = 2, the max is 2, meaning at least one die shows 2 and neither shows more than 2: (1,2), (2,1), (2,2), giving P(X = 2) = 3/16. For X = 3, we need max = 3: (1,3), (2,3), (3,1), (3,2), (3,3), giving P(X = 3) = 5/16. For X = 4, we need max = 4: (1,4), (2,4), (3,4), (4,1), (4,2), (4,3), (4,4), giving P(X = 4) = 7/16. The pattern 1, 3, 5, 7 emerges from counting systematically. A common misconception is assuming X-values 1, 2, 3, 4 are equally likely without considering how many pairs produce each maximum.
A fair six-sided die is rolled once (outcomes $1$ – $6$ equally likely). Define the random variable $X$ by the mapping: $X=0$ if the outcome is odd ( $1,3,5$ ) and $X=2$ if the outcome is even ( $2,4,6$ ). What is the expected value of $X$ based on this theoretical model?
Answer: $1$ This problem involves developing a theoretical probability distribution based on odd/even die outcomes. The model has 6 equally likely outcomes {1,2,3,4,5,6}, each with probability 1/6. The random variable X maps: X=0 when outcome is odd (1,3,5) and X=2 when outcome is even (2,4,6). To find P(X=x), we count outcomes: P(X=0) = P(1)+P(3)+P(5) = 3/6 = 1/2 (three odd numbers), and P(X=2) = P(2)+P(4)+P(6) = 3/6 = 1/2 (three even numbers). The expected value is E(X) = 0(1/2) + 2(1/2) = 0 + 1 = 1. A common misconception is forgetting to count all outcomes that map to each X-value, or assuming unequal probabilities for odd and even. The strategy: identify odd vs even outcomes → assign X-values → recognize equal counts give equal probabilities.
A fair six-sided die is rolled once (outcomes $1$ through $6$ are equally likely). Define the random variable $X$ as $X=1$ if the roll is prime ( $2,3,5$ ) and $X=0$ otherwise ( $1,4,6$ ). What is the expected value of $X$ based on this theoretical model?
Answer: $E(X)=\dfrac{1}{2}$ This problem involves creating a theoretical probability distribution for a prime indicator variable. The die has 6 equally likely outcomes, each with probability 1/6. We classify outcomes as prime (2,3,5) or not prime (1,4,6), mapping to X=1 and X=0 respectively. To find the distribution: P(X=1) = P(prime) = 3/6 = 1/2 (three prime outcomes), and P(X=0) = P(not prime) = 3/6 = 1/2 (three non-prime outcomes). The expected value is E(X) = 0(1/2) + 1(1/2) = 1/2. A misconception would be thinking we need to average the prime numbers themselves rather than the indicator values. The strategy: identify which outcomes satisfy the condition → assign indicator values → calculate probabilities.
A card is drawn at random from 6 equally likely cards labeled $A,B,C,D,E,F$ . Define the random variable $X$ by the mapping: $\{A\}\mapsto 0$ , $\{B,C\}\mapsto 1$ , and $\{D,E,F\}\mapsto 3$ . Which table correctly represents the theoretical probability distribution of $X$ ?
Answer: $P(X=0)=\frac{1}{6},\ P(X=1)=\frac{2}{6},\ P(X=3)=\frac{3}{6}$ This problem involves developing theoretical probability distributions and expected value, here focusing on the probability distribution. Probabilities are derived from the uniform model over six equally likely cards labeled A through F. The outcomes are mapped to X-values: A maps to 0, {B,C} to 1, and {D,E,F} to 3. The probability P(X=x) is determined by counting the cards mapping to each x and dividing by 6, yielding P(X=0)=1/6, P(X=1)=2/6, P(X=3)=3/6. This distribution can be used to compute expected value if needed, but here we identify the correct table. A common misconception is assuming all X-values are equally likely, but probabilities vary with the number of outcomes per value. To solve similar problems, list all outcomes, map each to X, and combine probabilities for each distinct X-value to build the distribution.
A fair six-sided die is rolled once (outcomes $1$ through $6$ are equally likely). Define the random variable $X$ as the number of factors of the rolled number. The mapping is: $1\mapsto 1$ , $2\mapsto 2$ , $3\mapsto 2$ , $4\mapsto 3$ , $5\mapsto 2$ , $6\mapsto 4$ . What is the expected value of $X$ based on this theoretical model?
Answer: $E(X)=\dfrac{7}{3}$ This problem maps die outcomes to their factor counts to find expected value. Each die outcome has probability 1/6. The factor counts are: 1→1 factor, 2→2 factors (1,2), 3→2 factors (1,3), 4→3 factors (1,2,4), 5→2 factors (1,5), 6→4 factors (1,2,3,6). The probability distribution is: P(X=1) = 1/6, P(X=2) = 3/6 = 1/2, P(X=3) = 1/6, P(X=4) = 1/6. The expected value is E(X) = 1(1/6) + 2(3/6) + 3(1/6) + 4(1/6) = 1/6 + 6/6 + 3/6 + 4/6 = 14/6 = 7/3. A misconception would be averaging the distinct values (1+2+3+4)/4 without considering that X=2 occurs for half the outcomes. The key insight: prime numbers 2,3,5 all have exactly 2 factors.
A fair six-sided die is rolled once (outcomes $1$ – $6$ equally likely). Define the random variable $X$ by the explicit mapping: $X=1$ if the outcome is $1,2,$ or $3$ ; $X=4$ if the outcome is $4$ or $5$ ; and $X=10$ if the outcome is $6$ . Which table correctly represents the theoretical probability distribution of $X$ ?
Answer: $P(X=1)=\dfrac{3}{6},\;P(X=4)=\dfrac{2}{6},\;P(X=10)=\dfrac{1}{6}$ This problem requires developing a theoretical probability distribution from an explicit mapping. The model has 6 equally likely die outcomes {1,2,3,4,5,6}, each with probability 1/6. The random variable X maps: X=1 when outcome is 1, 2, or 3; X=4 when outcome is 4 or 5; and X=10 when outcome is 6. To find P(X=x), we count how many outcomes map to each X-value: P(X=1) = P(1)+P(2)+P(3) = 3/6, P(X=4) = P(4)+P(5) = 2/6, and P(X=10) = P(6) = 1/6. The expected value would be E(X) = 1(3/6) + 4(2/6) + 10(1/6) = 3/6 + 8/6 + 10/6 = 21/6 = 7/2. A common misconception is thinking X-values {1,4,10} are equally likely with probability 1/3 each, ignoring the different number of die outcomes mapping to each. The strategy: identify which die outcomes produce each X-value → count them → express as fractions of 6.
A fair coin is flipped 3 times. The sample space consists of the 8 equally likely sequences (e.g., HHT, TTT). Define the random variable $X$ as the number of heads minus the number of tails in the 3 flips. (For example, HHT gives $X=1$ because $2-1=1$ .) Which table correctly represents the theoretical probability distribution of $X$ ?
Answer: $P(X=-3)=\dfrac{1}{8},\ P(X=-1)=\dfrac{3}{8},\ P(X=1)=\dfrac{3}{8},\ P(X=3)=\dfrac{1}{8}$ This problem involves finding the theoretical probability distribution of X = (number of heads) - (number of tails) when flipping a fair coin 3 times. The sample space contains 8 equally likely sequences, each with probability 1/8. To find X for each outcome, we count heads minus tails: HHH gives X = 3-0 = 3, HHT gives X = 2-1 = 1, HTH gives X = 2-1 = 1, HTT gives X = 1-2 = -1, THH gives X = 2-1 = 1, THT gives X = 1-2 = -1, TTH gives X = 1-2 = -1, and TTT gives X = 0-3 = -3. To find P(X = x), we count how many outcomes map to each x-value: P(X = -3) = 1/8 (only TTT), P(X = -1) = 3/8 (HTT, THT, TTH), P(X = 1) = 3/8 (HHT, HTH, THH), and P(X = 3) = 1/8 (only HHH). A common misconception is thinking the X-values themselves are equally likely, when in fact we must count how many coin sequences produce each X-value.

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