Statistics Quiz

Statistics Quiz: Developing Theoretical Probability Distributions Expected Value

Practice Developing Theoretical Probability Distributions Expected Value in Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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A fair die is rolled and a fair coin is flipped. The 12 outcomes in the sample space are equally likely. Define the random variable $X$ as follows: $X=1$ if the coin shows heads and the die shows a number greater than 4; otherwise $X=0$ . Which table correctly represents the theoretical probability distribution of $X$ ?

What this quiz covers

This quiz focuses on Developing Theoretical Probability Distributions Expected Value, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$P(X=0)=\tfrac{5}{12},\ P(X=1)=\tfrac{1}{12}$
$P(X=0)=\tfrac{5}{6},\ P(X=1)=\tfrac{1}{6}$ (correct answer)
$P(X=0)=\tfrac{5}{12},\ P(X=1)=\tfrac{7}{12}$
$P(X=0)=\tfrac{11}{12},\ P(X=1)=\tfrac{1}{12}$

Explanation: This problem involves developing theoretical probability distributions and calculating expected values for discrete random variables. In this model, probabilities are derived from the 12 equally likely outcomes of a fair die roll and coin flip, each with probability 1/12. Each outcome maps to X=1 if the coin is heads and die >4 (i.e., heads with 5 or 6), otherwise X=0. The probability P(X=x) is determined by counting favorable outcomes: P(X=1)=2/12=1/6, P(X=0)=10/12=5/6. The expected value E(X) is computed as 1*(1/6) + 0*(5/6) = 1/6. A common misconception is that all X-values are equally likely, but X=0 is far more probable due to the conditions. To solve similar problems, list all outcomes, map each to its X, and combine probabilities for each distinct x.

Question 2

A bag contains 4 equally likely cards labeled $1,2,3,4$ . One card is drawn at random. Define the random variable $X$ by the explicit mapping: $1\mapsto -1$ , $2\mapsto 0$ , $3\mapsto 0$ , $4\mapsto 2$ . Which table correctly represents the theoretical probability distribution of $X$ ?

$P(X=-1)=\tfrac{1}{3},\ P(X=0)=\tfrac{1}{3},\ P(X=2)=\tfrac{1}{3}$
$P(1)=\tfrac{1}{4},\ P(2)=\tfrac{1}{4},\ P(3)=\tfrac{1}{4},\ P(4)=\tfrac{1}{4}$
$P(X=-1)=\tfrac{1}{4},\ P(X=0)=\tfrac{1}{4},\ P(X=2)=\tfrac{1}{4}$
$P(X=-1)=\tfrac{1}{4},\ P(X=0)=\tfrac{1}{2},\ P(X=2)=\tfrac{1}{4}$ (correct answer)

Explanation: This problem involves developing theoretical probability distributions and calculating expected values for discrete random variables. In this model, probabilities are derived from the equally likely cards labeled 1 through 4, each with probability 1/4. Each card maps to an X-value via the explicit rule: 1 to -1, 2 to 0, 3 to 0, and 4 to 2. The probability P(X=x) is determined by summing probabilities of cards mapping to each x, resulting in P(X=-1)=1/4, P(X=0)=2/4=1/2, and P(X=2)=1/4. The expected value E(X) is computed as (-1)(1/4) + 0(1/2) + 2*(1/4) = (-1/4 + 0 + 2/4) = 1/4. A common misconception is that all X-values are equally likely, but P(X=0) is higher due to multiple mappings. To solve similar problems, list all outcomes, map each to its X, and combine probabilities for each distinct x.

Question 3

A fair coin is flipped 3 times. The sample space consists of all equally likely length-3 sequences of H and T. Define the random variable $X$ as the number of heads in the 3 flips. Which table correctly represents the theoretical probability distribution of $X$ (using $P(X=x)$ derived from the fraction of outcomes that produce each $x$ )?

$P(X=0)=\tfrac{1}{8},\ P(X=1)=\tfrac{3}{8},\ P(X=2)=\tfrac{3}{8},\ P(X=3)=\tfrac{1}{8}$ (correct answer)
$P(\text{HHH})=\tfrac{1}{8},\ P(\text{HHT})=\tfrac{1}{8},\ P(\text{HTH})=\tfrac{1}{8},\ P(\text{THH})=\tfrac{1}{8}$
$P(X=0)=\tfrac{1}{8},\ P(X=1)=\tfrac{2}{8},\ P(X=2)=\tfrac{4}{8},\ P(X=3)=\tfrac{1}{8}$
$P(X=0)=\tfrac{1}{4},\ P(X=1)=\tfrac{1}{4},\ P(X=2)=\tfrac{1}{4},\ P(X=3)=\tfrac{1}{4}$

Explanation: This problem involves developing theoretical probability distributions and calculating expected values for discrete random variables. In this model, probabilities are derived from the equally likely outcomes in the sample space of three fair coin flips, consisting of 8 sequences each with probability 1/8. Each outcome sequence maps to an X-value defined as the number of heads in the sequence. The probability P(X=x) is determined by counting the sequences with exactly x heads and dividing by 8, resulting in P(X=0)=1/8, P(X=1)=3/8, P(X=2)=3/8, and P(X=3)=1/8. The expected value E(X) is computed as the sum of x times P(X=x), yielding 0*(1/8) + 1*(3/8) + 2*(3/8) + 3*(1/8) = 12/8 = 3/2. A common misconception is that all X-values are equally likely, but X=1 and X=2 have higher probabilities due to more supporting outcomes. To solve similar problems, list all outcomes, map each to its X value, and combine probabilities for each distinct x.

Question 4

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 $X$ ?

$P(X=-1)=\tfrac{1}{2},\ P(X=0)=\tfrac{1}{4},\ P(X=2)=\tfrac{1}{4}$
$P(X=-1)=\tfrac{1}{4},\ P(X=0)=\tfrac{1}{2},\ P(X=2)=\tfrac{1}{4}$ (correct answer)
$P(X=-1)=\tfrac{1}{4},\ P(X=2)=\tfrac{1}{4}$
$P(X=-1)=\tfrac{1}{3},\ P(X=0)=\tfrac{1}{3},\ P(X=2)=\tfrac{1}{3}$

Explanation: 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.

Question 5

A spinner is divided into 5 equal sectors labeled $A,B,C,D,E$ , so each outcome is equally likely. Define the random variable $X$ by the mapping: $A\mapsto 0$ , $B\mapsto 0$ , $C\mapsto 1$ , $D\mapsto 2$ , $E\mapsto 5$ . What is the expected value of $X$ ?

$\tfrac{6}{5}$
$\tfrac{8}{5}$ (correct answer)
$\tfrac{8}{3}$
$\tfrac{5}{2}$

Explanation: This problem involves developing theoretical probability distributions and calculating expected values for discrete random variables. In this model, probabilities are derived from the five equally likely sectors of the spinner, each with probability 1/5. Each sector maps to an X-value: A and B to 0, C to 1, D to 2, and E to 5. The probability P(X=x) is determined by summing probabilities per x, yielding P(X=0)=2/5, P(X=1)=1/5, P(X=2)=1/5, P(X=5)=1/5. The expected value E(X) is computed as 0*(2/5) + 1*(1/5) + 2*(1/5) + 5*(1/5) = (0 + 1/5 + 2/5 + 5/5) = 8/5. A common misconception is assuming equal likelihood for all X-values, but P(X=0) is higher due to multiple sectors. To solve similar problems, list all outcomes, map each to its X, and combine probabilities before calculating E(X).

Question 6

Two fair coins are flipped; all 4 outcomes are equally likely: $\{\text{HH},\text{HT},\text{TH},\text{TT}\}$ . Define the random variable $X$ as follows: $X=2$ if both flips match (HH or TT) and $X=0$ otherwise (HT or TH). The mapping is $\{\text{HH},\text{TT}\}\to 2$ and $\{\text{HT},\text{TH}\}\to 0$ . What is the expected value of $X$ based on this theoretical model?

$2$
$1$ (correct answer)
$0$
$\frac{1}{2}$

Explanation: This scenario involves theoretical probability distributions and expected value for matching coin flips. Probabilities arise from the uniform model over four equally likely outcomes. Outcomes map to X=2 for matches {HH, TT} and X=0 for mismatches {HT, TH}. Thus, P(X=2) = 2/4 = 1/2 and P(X=0) = 2/4 = 1/2. The expected value is E[X] = 2*(1/2) + 0*(1/2) = 1. A common misconception is assuming all X-values equally likely without considering outcome counts, but here they are balanced. For similar problems, list all outcomes, map to X-categories, and combine probabilities accordingly.

Question 7

A fair coin is flipped 3 times; all $2^3=8$ outcomes are equally likely. Define the random variable $X$ as the number of heads. The outcomes map to $X$ as follows: $\text{TTT}\to 0$ ; $\text{HTT, THT, TTH}\to 1$ ; $\text{HHT, HTH, THH}\to 2$ ; $\text{HHH}\to 3$ . Which table correctly represents the theoretical probability distribution of $X$ ?

$P(X=0)=\frac{1}{8},\;P(X=1)=\frac{3}{8},\;P(X=2)=\frac{3}{8},\;P(X=3)=\frac{1}{8}$ (correct answer)
$P(X=0)=\frac{1}{4},\;P(X=1)=\frac{1}{4},\;P(X=2)=\frac{1}{4},\;P(X=3)=\frac{1}{4}$
$P(X=0)=\frac{1}{8},\;P(X=1)=\frac{1}{8},\;P(X=2)=\frac{3}{8},\;P(X=3)=\frac{3}{8}$
$P(\text{TTT})=\frac{1}{8},\;P(\text{HTT})=\frac{1}{8},\;P(\text{THT})=\frac{1}{8},\;P(\text{TTH})=\frac{1}{8}$

Explanation: This problem addresses theoretical probability distributions and expected value for the number of heads in coin flips. Probabilities are derived from the uniform model over 8 equally likely outcomes of three fair coin flips. Outcomes map to X-values based on head count: one outcome to X=0, three to X=1, three to X=2, and one to X=3. Thus, P(X=0) = 1/8, P(X=1) = 3/8, P(X=2) = 3/8, and P(X=3) = 1/8, matching choice A. This distribution justifies an expected value of E[X] = 0*(1/8) + 1*(3/8) + 2*(3/8) + 3*(1/8) = 1.5, though not asked. A misconception is assuming equal likelihood for each X-value, but probabilities vary with the number of supporting outcomes. For similar problems, list all outcomes, map to X, and combine probabilities by counting outcomes per X-value.

Question 8

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?

$\frac{1}{2}$
$2$
$\frac{2}{3}$
$1$ (correct answer)

Explanation: 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.

Question 9

A fair coin is flipped 4 times; all $16$ outcomes are equally likely. Define the random variable $X$ as the number of times two consecutive flips are both heads (counting overlaps). For example, $\text{HHHT}$ has two occurrences of consecutive heads (positions 1–2 and 2–3), so $X=2$ . The mapping by outcome type includes: $\text{TTTT}\to 0$ , $\text{HHTT}\to 1$ , $\text{HHHT}\to 2$ , $\text{HHHH}\to 3$ , etc. What is the expected value of $X$ based on this theoretical model?

$\frac{3}{4}$ (correct answer)
$1$
$\frac{1}{2}$
$\frac{3}{8}$

Explanation: This problem involves theoretical probability distributions and expected value for consecutive heads in coin flips. Probabilities arise from the uniform model over 16 equally likely outcomes of four fair coin flips. Outcomes map to X as the count of overlapping HH pairs in positions 1-2, 2-3, and 3-4. Using linearity, P(each pair HH) = 1/4, so E[X] = 3*(1/4) = 3/4, though the distribution is more complex due to dependencies. This confirms the expected value as 3/4. A misconception is assuming X-values are equally likely, but their probabilities reflect varying combinations. For related problems, list outcomes, map to X via indicators if needed, and combine for probabilities or directly compute expectations.

Question 10

A student guesses on a multiple-choice question with 4 answer choices (A–D). Exactly one choice is correct, and each choice is equally likely to be selected when guessing. Define the random variable $X$ as the score where $X=3$ points for a correct answer and $X=-1$ point for an incorrect answer. The mapping is: $\{\text{correct}\}\to 3$ and $\{\text{incorrect}\}\to -1$ (3 outcomes). Which table correctly represents the theoretical probability distribution of $X$ ?

$P(X=3)=\frac{1}{4},\;P(X=-1)=\frac{1}{4}$
$P(\text{A})=\frac{1}{4},\;P(\text{B})=\frac{1}{4},\;P(\text{C})=\frac{1}{4},\;P(\text{D})=\frac{1}{4}$
$P(X=3)=\frac{1}{4},\;P(X=-1)=\frac{3}{4}$ (correct answer)
$P(X=3)=\frac{1}{2},\;P(X=-1)=\frac{1}{2}$

Explanation: This question deals with theoretical probability distributions and expected value in guessing scenarios. Probabilities derive from the uniform model over four equally likely choices, with one correct. Outcomes map to X=3 for correct and X=-1 for each of the three incorrect. Thus, P(X=3) = 1/4 and P(X=-1) = 3/4, as in choice A. This gives E[X] = 3*(1/4) + (-1)*(3/4) = 0. A misconception is equal probability for the two X-values, but the incorrect has three times the support. For similar cases, list choices, map to scores, and combine probabilities for distinct X.

Question 11

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 $X$ ?

$P(X=0)=\tfrac13,\;P(X=1)=\tfrac13,\;P(X=2)=\tfrac13$
$P(X=0)=\tfrac14,\;P(X=1)=\tfrac14,\;P(X=2)=\tfrac12$
$P(HH)=\tfrac14,\;P(HT)=\tfrac14,\;P(TH)=\tfrac14,\;P(TT)=\tfrac14$
$P(X=0)=\tfrac14,\;P(X=1)=\tfrac12,\;P(X=2)=\tfrac14$ (correct answer)

Explanation: 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.

Question 12

A fair four-sided die has equally likely outcomes $\{1,2,3,4\}$ . Define the random variable $X$ as $X=|\text{roll}-2|$ . The mapping is: $1\mapsto 1$ , $2\mapsto 0$ , $3\mapsto 1$ , $4\mapsto 2$ . Which table correctly represents the theoretical probability distribution of $X$ ?

$P(X=0)=\tfrac14,\;P(X=1)=\tfrac14,\;P(X=2)=\tfrac12$
$P(X=0)=\tfrac14,\;P(X=1)=\tfrac12$
$P(X=0)=\tfrac14,\;P(X=1)=\tfrac12,\;P(X=2)=\tfrac14$ (correct answer)
$P(X=0)=\tfrac13,\;P(X=1)=\tfrac13,\;P(X=2)=\tfrac13$

Explanation: This problem requires mapping die outcomes through an absolute value function to create a probability distribution. The four equally likely outcomes {1,2,3,4} each have probability 1/4. Applying X = |roll-2|: 1→|1-2|=1, 2→|2-2|=0, 3→|3-2|=1, 4→|4-2|=2. To find P(X=x), we combine probabilities of outcomes mapping to the same X-value: P(X=0) = P(roll=2) = 1/4, P(X=1) = P(roll=1) + P(roll=3) = 1/4 + 1/4 = 1/2, P(X=2) = P(roll=4) = 1/4. Notice that X=1 occurs twice as often as other values because two different rolls produce it. The transfer strategy: apply the function to each outcome → group by resulting values → sum probabilities.

Question 13

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?

$E(X)=\dfrac{1+2+2+3+2+4}{6}=\dfrac{7}{3}$
$E(X)=2$
$E(X)=\dfrac{1+2+3+4}{4}=\dfrac{5}{2}$
$E(X)=\dfrac{7}{3}$ (correct answer)

Explanation: 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.

Question 14

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 $X$ ?

$P(X=0)=\tfrac18,\;P(X=1)=\tfrac18,\;P(X=2)=\tfrac38,\;P(X=3)=\tfrac38$
$P(X=0)=\tfrac18,\;P(X=1)=\tfrac38,\;P(X=2)=\tfrac38,\;P(X=3)=\tfrac18$ (correct answer)
$P(X=0)=\tfrac18,\;P(X=1)=\tfrac38,\;P(X=2)=\tfrac18,\;P(X=3)=\tfrac38$
$P(X=0)=\tfrac14,\;P(X=1)=\tfrac14,\;P(X=2)=\tfrac14,\;P(X=3)=\tfrac14$

Explanation: 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).

Question 15

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?

$E(X)=\dfrac{8}{3}$
$E(X)=\dfrac{-2+0+3+9}{4}=\dfrac{5}{2}$
$E(X)=\dfrac{4}{3}$ (correct answer)
$E(X)=\dfrac{5}{3}$

Explanation: 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.

Question 16

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?

$E(X)=\dfrac{2}{5}$ (correct answer)
$E(X)=\dfrac{6}{5}$
$E(X)=\dfrac{4}{5}$
$E(X)=1$

Explanation: 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.

Question 17

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)?

$P(X=3)=\frac{1}{8},\ P(X=1)=\frac{3}{8},\ P(X=-1)=\frac{3}{8},\ P(X=-3)=\frac{1}{8}$ (correct answer)
$P(\text{HHH})=\frac{1}{8},\ P(\text{HHT})=\frac{1}{8},\ P(\text{HTH})=\frac{1}{8},\ P(\text{THH})=\frac{1}{8}$
$P(X=3)=\frac{1}{4},\ P(X=1)=\frac{1}{4},\ P(X=-1)=\frac{1}{4},\ P(X=-3)=\frac{1}{4}$
$P(X=3)=\frac{1}{8},\ P(X=1)=\frac{3}{8},\ P(X=-1)=\frac{1}{8}$

Explanation: 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.

Question 18

Two fair six-sided dice are rolled. The sample space has $36$ equally likely ordered outcomes $(a,b)$ . Define the random variable $X$ as the sum $a+b$ .

Which table correctly represents the theoretical probability distribution of $X$ ?

$P(X=2)=\frac{1}{36},\ P(X=3)=\frac{2}{36},\ P(X=4)=\frac{3}{36},\ P(X=5)=\frac{4}{36},\ P(X=6)=\frac{5}{36},\ P(X=7)=\frac{6}{36},\ P(X=8)=\frac{5}{36},\ P(X=9)=\frac{4}{36},\ P(X=10)=\frac{3}{36},\ P(X=11)=\frac{2}{36},\ P(X=12)=\frac{1}{36}$ (correct answer)
$P(X=2)=\frac{1}{11},\ P(X=3)=\frac{1}{11},\ldots,\ P(X=12)=\frac{1}{11}$
$P((1,1))=\frac{1}{36},\ P((1,2))=\frac{1}{36},\ P((1,3))=\frac{1}{36},\ P((1,4))=\frac{1}{36}$
$P(X=2)=\frac{1}{36},\ P(X=3)=\frac{2}{36},\ P(X=4)=\frac{3}{36},\ P(X=5)=\frac{4}{36},\ P(X=6)=\frac{5}{36},\ P(X=7)=\frac{6}{36},\ P(X=8)=\frac{6}{36},\ P(X=9)=\frac{5}{36},\ P(X=10)=\frac{4}{36},\ P(X=11)=\frac{3}{36},\ P(X=12)=\frac{2}{36}$

Explanation: This problem involves developing theoretical probability distributions and expected value, here focusing on the probability distribution. Probabilities are derived from the uniform model over 36 equally likely ordered pairs for two fair dice. The outcomes are mapped to X-values as the sum of the two dice. The probability P(X=x) is determined by counting the pairs summing to each x and dividing by 36, resulting in the asymmetric distribution peaking at 7. The correct table lists P(X=2)=1/36, P(X=3)=2/36, up to P(X=7)=6/36, then decreasing symmetrically to P(X=12)=1/36. A common misconception is assuming all sums from 2 to 12 are equally likely, but central sums have more combinations. To solve similar problems, list all outcomes, map each to X, and combine probabilities for each distinct X-value to build the distribution.

Question 19

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 $X$ ?

$P(X=1)=\dfrac{1}{16},\ P(X=2)=\dfrac{3}{16},\ P(X=3)=\dfrac{5}{16},\ P(X=4)=\dfrac{7}{16}$ (correct answer)
$P(X=1)=\dfrac{1}{4},\ P(X=2)=\dfrac{1}{4},\ P(X=3)=\dfrac{1}{4},\ P(X=4)=\dfrac{1}{4}$
$P(X=1)=\dfrac{1}{16},\ P(X=2)=\dfrac{2}{16},\ P(X=3)=\dfrac{3}{16},\ P(X=4)=\dfrac{4}{16}$
$P(X=1)=\dfrac{1}{16},\ P(X=2)=\dfrac{5}{16},\ P(X=3)=\dfrac{7}{16},\ P(X=4)=\dfrac{3}{16}$

Explanation: 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.

Question 20

A bag contains 3 red balls (R) and 2 blue balls (B). Two balls are drawn without replacement. All $\binom{5}{2}=10$ unordered pairs of balls are equally likely. Define the random variable $X$ as the number of red balls drawn. Which table correctly represents the theoretical probability distribution of $X$ ?

$P(X=0)=\dfrac{1}{10},\ P(X=1)=\dfrac{6}{10},\ P(X=2)=\dfrac{3}{10}$ (correct answer)
$P(X=0)=\dfrac{4}{10},\ P(X=1)=\dfrac{3}{10},\ P(X=2)=\dfrac{3}{10}$
$P(X=0)=\dfrac{1}{5},\ P(X=1)=\dfrac{3}{5},\ P(X=2)=\dfrac{1}{5}$
$P(X=0)=\dfrac{1}{10},\ P(X=2)=\dfrac{3}{10}$

Explanation: This problem involves finding the theoretical probability distribution of X = number of red balls when drawing 2 balls without replacement from a bag with 3 red and 2 blue balls. The sample space consists of all C(5,2) = 10 unordered pairs, each equally likely with probability 1/10. To find P(X = 0), we count pairs with no red balls: only {B,B}, which is C(2,2) = 1 way, so P(X = 0) = 1/10. For P(X = 1), we need exactly one red and one blue: choose 1 red from 3 and 1 blue from 2, giving C(3,1)×C(2,1) = 3×2 = 6 ways, so P(X = 1) = 6/10. For P(X = 2), we need two red balls: choose 2 from 3 red balls, giving C(3,2) = 3 ways, so P(X = 2) = 3/10. The distribution is P(X = 0) = 1/10, P(X = 1) = 6/10, P(X = 2) = 3/10. A common misconception is thinking X-values 0, 1, 2 are equally likely without using combinatorial counting.

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