Statistics Quiz: Defining Random Variables And Probability Distributions

20 questions · exam conditions

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Defining Random Variables And Probability DistributionsQuestion 1 of 20

A fair coin is flipped twice. The sample space is $\{HH, HT, TH, TT\}$ . Define the random variable $X$ as the number of heads obtained. Which table correctly represents the probability distribution of $X$ ?

$X$	$P(X)$
0	$\tfrac{1}{4}$
1	$\tfrac{1}{2}$
2	$\tfrac{1}{2}$

$X$	$P(X)$
0	$\tfrac{1}{4}$
1	$\tfrac{2}{4}$
2	$\tfrac{1}{4}$

$X$	$P(X)$
HH	$\tfrac{1}{4}$
HT	$\tfrac{1}{4}$
TH	$\tfrac{1}{4}$
TT	$\tfrac{1}{4}$

$X$	$P(X)$
0	$\tfrac{1}{4}$
2	$\tfrac{1}{4}$

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Statistics Quiz: Defining Random Variables And Probability Distributions

Practice Defining Random Variables And Probability Distributions in Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Defining Random Variables And Probability Distributions, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.

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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

$X$ $P(X)$
0 $\tfrac{1}{4}$
1 $\tfrac{1}{2}$
2 $\tfrac{1}{2}$
$X$ $P(X)$
0 $\tfrac{1}{4}$
1 $\tfrac{2}{4}$
2 $\tfrac{1}{4}$
(correct answer)
$X$ $P(X)$
HH $\tfrac{1}{4}$
HT $\tfrac{1}{4}$
TH $\tfrac{1}{4}$
TT $\tfrac{1}{4}$
$X$ $P(X)$
0 $\tfrac{1}{4}$
2 $\tfrac{1}{4}$

Explanation: Random variables and their probability distributions are key concepts in statistics, where a random variable X assigns a numerical value to each outcome in a sample space. Here, X counts the number of heads in two flips of a fair coin, with sample space {HH, HT, TH, TT}. Each outcome maps to X as follows: HH to 2, HT and TH to 1, TT to 0. Probabilities are assigned by counting equally likely outcomes: P(X=0) = 1/4, P(X=1) = 2/4, P(X=2) = 1/4. This distribution is complete and valid as it covers all possible X values (0, 1, 2) and sums to 1. A common misconception is listing the sample space as the distribution instead of grouping by X values; the distribution focuses on X, not raw outcomes. To build it, list outcomes, map to X, and sum probabilities for each X.

Question 2

A spinner is divided into 4 equal sections labeled 1, 1, 2, and 3. The spinner is spun once. Define the random variable $X$ as the number shown on the spinner. Which table correctly represents the probability distribution of $X$ ?

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

$X$	$P(X)$
1	$\tfrac{2}{4}$
2	$\tfrac{1}{4}$
3	$\tfrac{1}{4}$

$X$	$P(X)$
1	$\tfrac{2}{4}$
2	$\tfrac{2}{4}$
3	$\tfrac{1}{4}$

$X$	$P(X)$
1	$\tfrac{1}{4}$
2	$\tfrac{1}{4}$
3	$\tfrac{1}{4}$

Explanation: Random variables and their probability distributions are key concepts in statistics, where a random variable X assigns a numerical value to each outcome in a sample space. For a spinner with sections 1, 1, 2, 3, X is the number shown, so outcomes map directly to X=1, 2, or 3. Since two sections are 1, it maps to X=1 with higher probability. Probabilities are P(X=1) = 2/4, P(X=2) = 1/4, P(X=3) = 1/4, reflecting equal section chances. The distribution is complete and valid, including all X values (1, 2, 3) with probabilities summing to 1. A misconception is treating all X values as equally likely when outcomes aren't; here, X=1 combines two outcomes. Strategy: list outcomes, map to X, combine probabilities for duplicates.

Question 3

A box contains 3 tickets labeled $-1$ , 0, and 2. One ticket is drawn at random. Define the random variable $X$ as the value on the ticket drawn. Which table correctly represents the probability distribution of $X$ ?

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

$X$	$P(X)$
-1	$\tfrac{1}{3}$
2	$\tfrac{2}{3}$

$X$	$P(X)$
-1	$\tfrac{1}{2}$
0	$\tfrac{1}{3}$
2	$\tfrac{1}{3}$

$X$	$P(X)$
-1	$\tfrac{1}{3}$
0	$\tfrac{1}{3}$
2	$\tfrac{1}{3}$

$X$	$P(X)$
-1	$\tfrac{1}{3}$
0	$\tfrac{1}{3}$
1	$\tfrac{1}{3}$

Explanation: Random variables and their probability distributions are key concepts in statistics, where a random variable X assigns a numerical value to each outcome in a sample space. With tickets -1,0,2, X is the value drawn, mapping each ticket directly to its X. The three outcomes map to distinct X=-1,0,2. Since equally likely, each P(X) = 1/3. The distribution is complete and valid, including all X with probabilities summing to 1. Misconception: assuming X values must be non-negative; they can be negative like here. Process: list tickets, map to X, assign equal probabilities.

Question 4

A fair six-sided die is rolled once. Define the random variable $X$ as the number of even outcomes rolled. Which table correctly represents the probability distribution of $X$ ?

$X$ $P(X)$
2 $\tfrac{1}{2}$
4 $\tfrac{1}{2}$
$X$ $P(X)$
0 $\tfrac{1}{2}$
1 $\tfrac{1}{2}$
(correct answer)
$X$ $P(X)$
0 $\tfrac{1}{6}$
1 $\tfrac{5}{6}$
$X$ $P(X)$
0 $\tfrac{3}{6}$
1 $\tfrac{2}{6}$

$X$	$P(X)$
2	$\tfrac{1}{2}$
4	$\tfrac{1}{2}$

$X$	$P(X)$
0	$\tfrac{1}{2}$
1	$\tfrac{1}{2}$

$X$	$P(X)$
0	$\tfrac{1}{6}$
1	$\tfrac{5}{6}$

$X$	$P(X)$
0	$\tfrac{3}{6}$
1	$\tfrac{2}{6}$

Explanation: Random variables and their probability distributions are key concepts in statistics, where a random variable X assigns a numerical value to each outcome in a sample space. In a die roll, X is the count of even numbers, so odd outcomes map to X=0 and even to X=1. The six faces map as: 1,3,5 to 0; 2,4,6 to 1. Probabilities are P(X=0) = 3/6 = 1/2 and P(X=1) = 3/6 = 1/2, since faces are equally likely. This is complete and valid, covering X=0 and 1 with sum 1. Misconception: thinking X counts all evens possible instead of per roll; it's binary here. Approach: list outcomes, map to X, aggregate probabilities.

Question 5

A bag contains 2 red marbles (R) and 1 blue marble (B). One marble is drawn at random. Define the random variable $X$ as the number of red marbles drawn. Which table correctly represents the probability distribution of $X$ ?

$X$ $P(X)$
0 $\tfrac{1}{3}$
1 $\tfrac{2}{3}$
(correct answer)
$X$ $P(X)$
0 $\tfrac{2}{3}$
1 $\tfrac{1}{3}$
$X$ $P(X)$
0 $\tfrac{1}{3}$
1 $\tfrac{2}{3}$
2 $\tfrac{1}{3}$
$X$ $P(X)$
R $\tfrac{2}{3}$
B $\tfrac{1}{3}$

$X$	$P(X)$
0	$\tfrac{1}{3}$
1	$\tfrac{2}{3}$

$X$	$P(X)$
0	$\tfrac{2}{3}$
1	$\tfrac{1}{3}$

$X$	$P(X)$
0	$\tfrac{1}{3}$
1	$\tfrac{2}{3}$
2	$\tfrac{1}{3}$

$X$	$P(X)$
R	$\tfrac{2}{3}$
B	$\tfrac{1}{3}$

Explanation: Random variables and their probability distributions are key concepts in statistics, where a random variable X assigns a numerical value to each outcome in a sample space. In this case, X represents the number of red marbles drawn from a bag with 2 red and 1 blue marble. The possible outcomes are drawing a red (X=1) or blue (X=0) marble. The probabilities are assigned based on the proportions: P(X=1) = 2/3 for red and P(X=0) = 1/3 for blue. This distribution is complete and valid because it includes all possible values of X (0 and 1) and the probabilities sum to 1. A common misconception is confusing the sample space outcomes (R, B) with the values of X; here, multiple outcomes map to the same X only if there were more draws, but it's a single draw. To find such distributions, list all outcomes, map each to its X value, and combine probabilities for each unique X.

Question 6

Two cards are drawn without replacement from a standard deck of 52 cards. Define the random variable $X$ as the number of aces drawn. Which table correctly represents the probability distribution of $X$ ?

$X$ $P(X)$
0 $\tfrac{48}{52}\cdot\tfrac{47}{51}$
1 $\tfrac{4}{52}\cdot\tfrac{48}{51}$
2 $\tfrac{4}{52}\cdot\tfrac{3}{51}$
$X$ $P(X)$
0 $\tfrac{48}{52}\cdot\tfrac{47}{51}$
2 $\tfrac{4}{52}\cdot\tfrac{3}{51}$

$X$	$P(X)$
0	$\tfrac{48}{52}\cdot\tfrac{47}{51}$
1	$\tfrac{4}{52}\cdot\tfrac{48}{51}+\tfrac{48}{52}\cdot\tfrac{4}{51}$
2	$\tfrac{4}{52}\cdot\tfrac{3}{51}$

(correct answer)

$X$ $P(X)$
0 $\tfrac{48}{52}$
1 $\tfrac{4}{52}$
2 $\tfrac{4}{52}\cdot\tfrac{3}{51}$

$X$	$P(X)$
0	$\tfrac{48}{52}$
1	$\tfrac{4}{52}$
2	$\tfrac{4}{52}\cdot\tfrac{3}{51}$

Explanation: Random variables and their probability distributions are key concepts in statistics, where a random variable X assigns a numerical value to each outcome in a sample space. For drawing two cards, X counts aces, so outcomes map to X=0,1, or 2 based on aces drawn. Paths like ace then non-ace or vice versa both map to X=1. Probabilities account for order: P(X=0) = (48/52)(47/51), P(X=1) = (4/52)(48/51) + (48/52)(4/51), P(X=2) = (4/52)(3/51). It's complete and valid, covering all X with sum 1. Misconception: ignoring order in without-replacement draws; both sequences for X=1 must combine. Technique: list draw sequences, map to X, add probabilities.

Question 7

A survey asks a randomly selected student how many siblings they have. Suppose the only possible responses in this group are 0, 1, or 2 siblings, with probabilities $P(0)=0.30$ , $P(1)=0.50$ , and $P(2)=0.20$ . Let $X$ be the number of siblings reported. Which graph represents the distribution of $X$ ?

A bar graph with bars at $X=0,1,2$ having heights $0.30, 0.50, 0.20$ respectively. (correct answer)
A bar graph with bars at $X=0,1,2,3$ having heights $0.30, 0.50, 0.20, 0.10$ respectively.
A bar graph with bars at $X=0,1$ having heights $0.30, 0.50$ (no bar at $X=2$ ).
A bar graph with bars at $X=0,1,2$ having heights $0.20, 0.50, 0.30$ respectively.

Explanation: Here, the concept is random variables and distributions represented graphically for sibling counts. A random variable assigns numbers to survey responses, with X as the number of siblings (0,1, or 2). Each response maps directly to X=0,1, or 2 with given probabilities 0.30, 0.50, 0.20. These probabilities are assigned to X values and visualized as bar heights. The graph in choice A is complete and valid, showing bars for all possible X (0,1,2) with correct heights summing to 1 implicitly via the probabilities. Misconception: confusing response categories with X values, but X is the numerical count. To create this, list possible responses, map to X, and plot combined probabilities as bar heights.

Question 8

A spinner has 4 equal sections labeled A, B, C, and D. You spin once. Define the random variable $X$ by: $X=2$ if the result is A or B, $X=5$ if the result is C, and $X=8$ if the result is D. Which table correctly represents the probability distribution of $X$ ?

$X$ $P(X)$
A $\tfrac{1}{4}$
B $\tfrac{1}{4}$
C $\tfrac{1}{4}$
D $\tfrac{1}{4}$
$X$ $P(X)$
2 $\tfrac{2}{5}$
5 $\tfrac{1}{5}$
8 $\tfrac{2}{5}$
$X$ $P(X)$
2 $\tfrac{1}{2}$
5 $\tfrac{1}{4}$
8 $\tfrac{1}{4}$
(correct answer)
$X$ $P(X)$
2 $\tfrac{1}{4}$
5 $\tfrac{1}{4}$
8 $\tfrac{1}{4}$

$X$	$P(X)$
A	$\tfrac{1}{4}$
B	$\tfrac{1}{4}$
C	$\tfrac{1}{4}$
D	$\tfrac{1}{4}$

$X$	$P(X)$
2	$\tfrac{2}{5}$
5	$\tfrac{1}{5}$
8	$\tfrac{2}{5}$

$X$	$P(X)$
2	$\tfrac{1}{2}$
5	$\tfrac{1}{4}$
8	$\tfrac{1}{4}$

$X$	$P(X)$
2	$\tfrac{1}{4}$
5	$\tfrac{1}{4}$
8	$\tfrac{1}{4}$

Explanation: The key concept is defining random variables and distributions based on spinner outcomes mapped to custom values. A random variable assigns numerical values to outcomes, with X=2 for A or B, X=5 for C, and X=8 for D on a four-section spinner. Each spin outcome (A, B, C, D) maps to its corresponding X value, with equal probability 1/4 per section. Probabilities for X are aggregated: P(X=2)=P(A)+P(B)=1/2, P(X=5)=1/4, P(X=8)=1/4. The distribution in choice C is complete and valid because it covers all possible X (2,5,8), sums to 1, and correctly combines probabilities from multiple outcomes to each X. People often confuse outcomes like 'A' with X values, but X is the assigned number. To build this, list spinner outcomes, map to X, and sum probabilities for duplicate X values.

Question 9

A box contains 1 gold ticket and 2 silver tickets. One ticket is drawn at random. You win $10$ for a gold ticket and $4$ for a silver ticket. Let $X$ be the amount of money won. Which table correctly represents the probability distribution of $X$ ?

$X$ $P(X)$
gold $\tfrac{1}{3}$
silver $\tfrac{2}{3}$
$X$ $P(X)$
$4$ $\tfrac{1}{3}$
$10$ $\tfrac{2}{3}$
$X$ $P(X)$
$4$ $\tfrac{2}{3}$
$10$ $\tfrac{2}{3}$
$X$ $P(X)$
$10$ $\tfrac{1}{3}$
(correct answer)

$X$	$P(X)$
gold	$\tfrac{1}{3}$
silver	$\tfrac{2}{3}$

$X$	$P(X)$
$4$	$\tfrac{1}{3}$
$10$	$\tfrac{2}{3}$

$X$	$P(X)$
$4$	$\tfrac{2}{3}$
$10$	$\tfrac{2}{3}$

$X$	$P(X)$
$10$	$\tfrac{1}{3}$

Explanation: This example covers random variables and distributions for winnings from ticket draws. A random variable assigns numbers to outcomes, with

X=10

for gold,

X=4

for silver, from one gold and two silver tickets. Each ticket maps to its

X

value, with

P

(gold)=\frac{1}{3}

,

(silver)=\frac{2}{3}

. Thus,

P(X=10)=\frac{1}{3}

P(X=4)=\frac{2}{3}

. Choice A's distribution is complete and valid, listing distinct

X

(

4

10

) with probabilities summing to 1 and matching the draw odds. Misconception: using 'gold' or 'silver' as

X

instead of monetary values. To derive: list three tickets, map to

X

, combine probabilities for same

X

(though none here).

Question 10

A fair six-sided die is rolled once. Define the random variable $X$ as the number of points you earn, where you earn 0 points if you roll 1 or 2, 1 point if you roll 3 or 4, and 2 points if you roll 5 or 6. Which table correctly represents the probability distribution of $X$ ?

$X$ $P(X)$
0 $\tfrac{2}{6}$
1 $\tfrac{2}{6}$
2 $\tfrac{1}{6}$
$X$ $P(X)$
0 $\tfrac{2}{6}$
1 $\tfrac{2}{6}$
2 $\tfrac{2}{6}$
(correct answer)
$X$ $P(X)$
0 $\tfrac{2}{6}$
1 $\tfrac{2}{6}$
$X$ $P(X)$
0 $\tfrac{1}{6}$
1 $\tfrac{2}{6}$
2 $\tfrac{3}{6}$

$X$	$P(X)$
0	$\tfrac{2}{6}$
1	$\tfrac{2}{6}$
2	$\tfrac{1}{6}$

$X$	$P(X)$
0	$\tfrac{2}{6}$
1	$\tfrac{2}{6}$
2	$\tfrac{2}{6}$

$X$	$P(X)$
0	$\tfrac{2}{6}$
1	$\tfrac{2}{6}$

$X$	$P(X)$
0	$\tfrac{1}{6}$
1	$\tfrac{2}{6}$
2	$\tfrac{3}{6}$

Explanation: This example demonstrates random variables and distributions for points earned from a die roll. A random variable assigns numbers to outcomes, here X=0 for rolls 1-2, X=1 for 3-4, X=2 for 5-6. Each die face maps to an X value, with two faces per category. Since each face has probability 1/6, P(X=0)=2/6, P(X=1)=2/6, P(X=2)=2/6. The distribution in choice A is complete and valid as it includes all possible X (0,1,2), probabilities sum to 1, and it accurately groups the equal probabilities. A misconception is equating die faces directly to X without grouping. The approach: list all six outcomes, map to X, and combine probabilities for each X.

Question 11

A jar contains 2 green marbles and 2 yellow marbles. You draw 2 marbles without replacement. Define the random variable $X$ as the number of green marbles drawn. Which table correctly represents the probability distribution of $X$ ?

$X$ $P(X)$
0 $\tfrac{1}{6}$
1 $\tfrac{4}{6}$
2 $\tfrac{1}{6}$
(correct answer)
$X$ $P(X)$
0 $\tfrac{1}{4}$
1 $\tfrac{1}{2}$
2 $\tfrac{1}{4}$
$X$ $P(X)$
0 $\tfrac{1}{6}$
2 $\tfrac{1}{6}$
$X$ $P(X)$
0 $\tfrac{1}{6}$
1 $\tfrac{2}{6}$
2 $\tfrac{1}{6}$

$X$	$P(X)$
0	$\tfrac{1}{6}$
1	$\tfrac{4}{6}$
2	$\tfrac{1}{6}$

$X$	$P(X)$
0	$\tfrac{1}{6}$
2	$\tfrac{1}{6}$

$X$	$P(X)$
0	$\tfrac{1}{6}$
1	$\tfrac{2}{6}$
2	$\tfrac{1}{6}$

Explanation: This question explores random variables and distributions for marble draws without replacement. A random variable assigns numbers to outcomes, with X as the count of green marbles in a draw of two from two green and two yellow. Possible outcomes are pairs like GG, GY, YY, mapping to X=2,1,0 respectively. Using combinations, P(X=0)=C(2,2)/C(4,2)=1/6, P(X=1)=[C(2,1)*C(2,1)]/6=4/6, P(X=2)=1/6. Choice A's distribution is complete and valid, covering X=0,1,2 with probabilities summing to 1 and reflecting the hypergeometric setup. Misconception: listing pairs like GY as X instead of the green count. Method: list all six pairs, map to X, combine probabilities.

Question 12

A jar contains 2 green balls and 2 yellow balls. Two balls are drawn without replacement. Let $X$ be the number of green balls drawn. Which table correctly represents the probability distribution of $X$ ?

$\begin{array}{c|c}X & P(X)\\\hline 0 & \tfrac{1}{4}\\ 1 & \tfrac{1}{2}\\ 2 & \tfrac{1}{4}\end{array}$
$\begin{array}{c|c}X & P(X)\\\hline 0 & \tfrac{1}{6}\\ 1 & \tfrac{2}{3}\end{array}$
$\begin{array}{c|c}X & P(X)\\\hline 0 & \tfrac{1}{6}\\ 1 & \tfrac{2}{3}\\ 2 & \tfrac{1}{3}\end{array}$
$\begin{array}{c|c}X & P(X)\\\hline 0 & \tfrac{1}{6}\\ 1 & \tfrac{2}{3}\\ 2 & \tfrac{1}{6}\end{array}$ (correct answer)

Explanation: This problem involves drawing 2 balls without replacement from 2 green and 2 yellow balls, with X counting green balls drawn. A random variable X assigns numbers to outcomes - here it counts green balls in each draw combination. The possible outcomes and their X-values are: GG → X = 2, GY → X = 1, YG → X = 1, YY → X = 0. For probabilities without replacement: P(GG) = (2/4)(1/3) = 1/6, P(GY) = (2/4)(2/3) = 1/3, P(YG) = (2/4)(2/3) = 1/3, P(YY) = (2/4)(1/3) = 1/6. Combining same X-values: P(X = 0) = 1/6, P(X = 1) = 1/3 + 1/3 = 2/3, P(X = 2) = 1/6. Option A correctly shows this complete distribution. The key insight: without replacement changes probabilities for the second draw based on the first.

Question 13

A survey randomly selects one student from a class. The student's grade level is equally likely to be 9th, 10th, 11th, or 12th. Define the random variable $X$ as the grade level number (9, 10, 11, or 12). Which table correctly represents the probability distribution of $X$ ?

$\begin{array}{c|c}X & P(X)\\\hline 9 & 0.25\\ 10 & 0.25\\ 11 & 0.25\end{array}$
$\begin{array}{c|c}X & P(X)\\\hline \text{9th} & 0.25\\ \text{10th} & 0.25\\ \text{11th} & 0.25\\ \text{12th} & 0.25\end{array}$
$\begin{array}{c|c}X & P(X)\\\hline 9 & 0.25\\ 10 & 0.25\\ 11 & 0.25\\ 12 & 0.25\end{array}$ (correct answer)
$\begin{array}{c|c}X & P(X)\\\hline 9 & 0.30\\ 10 & 0.30\\ 11 & 0.30\\ 12 & 0.30\end{array}$

Explanation: This problem involves selecting a student where X represents their grade level number. A random variable X assigns numerical values to outcomes - here X equals the grade level (9, 10, 11, or 12). Since each grade level is equally likely, the four outcomes have equal probability. With 4 possible outcomes, each has probability 1/4 = 0.25: P(X = 9) = P(X = 10) = P(X = 11) = P(X = 12) = 0.25. Option A correctly shows all four X-values with equal probabilities that sum to 1. Option D incorrectly uses ordinal labels (9th, 10th) instead of the numerical X-values - the random variable X is defined as the grade number itself. To verify: list all possible X-values → assign equal probabilities → ensure they sum to 1.

Question 14

A game is played by drawing one card at random from a set of 5 cards labeled $\{1,1,2,3,3\}$ . You win an amount equal to the number on the card. Let $X$ be your winnings (in dollars). Which table correctly represents the probability distribution of $X$ ?

$\begin{array}{c|c}X & P(X)\\\hline 1 & \tfrac{2}{5}\\ 2 & \tfrac{1}{5}\end{array}$
$\begin{array}{c|c}X & P(X)\\\hline 1 & \tfrac{2}{5}\\ 2 & \tfrac{1}{5}\\ 3 & \tfrac{2}{5}\end{array}$ (correct answer)
$\begin{array}{c|c}X & P(X)\\\hline 1 & \tfrac{1}{5}\\ 2 & \tfrac{1}{5}\\ 3 & \tfrac{1}{5}\end{array}$
$\begin{array}{c|c}X & P(X)\\\hline 1 & \tfrac{2}{5}\\ 2 & \tfrac{1}{5}\\ 3 & \tfrac{1}{5}\end{array}$

Explanation: This problem asks for the probability distribution of winnings

X

when drawing from cards

\{1, 1, 2, 3, 3\}

. A random variable

X

represents the dollar amount won, which equals the card value drawn. The possible outcomes are drawing cards labeled 1, 2, or 3, giving

X

-values

\{1, 2, 3\}

. To find probabilities:

P(X = 1) = \frac{2}{5}

(two 1's out of 5 cards),

P(X = 2) = \frac{1}{5}

(one 2 out of 5 cards),

P(X = 3) = \frac{2}{5}

(two 3's out of 5 cards). Option A correctly shows this distribution with all

X

-values and their probabilities summing to 1. Option C omits

X = 3

, making it incomplete - a probability distribution must include all possible values of

X

. The strategy: count how many cards give each

X

-value → divide by total cards → list all

X

-values with probabilities.

Question 15

A bag contains 3 red marbles and 1 blue marble. One marble is drawn at random, its color is recorded, and then it is returned to the bag. This is repeated for a total of 2 draws. Let $X$ be the number of red marbles drawn in the 2 draws. Which table correctly represents the probability distribution of $X$ ?

$X$ $P(X)$
0 $\frac{1}{16}$
1 $\frac{6}{16}$
2 $\frac{9}{16}$
(correct answer)
$X$ $P(X)$
0 $\frac{1}{16}$
2 $\frac{9}{16}$
$X$ $P(X)$
0 $\frac{1}{4}$
1 $\frac{1}{2}$
2 $\frac{1}{4}$
Outcome Probability
RR $\frac{9}{16}$
RB $\frac{3}{16}$
BR $\frac{3}{16}$
BB $\frac{1}{16}$

$X$	$P(X)$
0	$\frac{1}{16}$
1	$\frac{6}{16}$
2	$\frac{9}{16}$

$X$	$P(X)$
0	$\frac{1}{16}$
2	$\frac{9}{16}$

$X$	$P(X)$
0	$\frac{1}{4}$
1	$\frac{1}{2}$
2	$\frac{1}{4}$

Outcome	Probability
RR	$\frac{9}{16}$
RB	$\frac{3}{16}$
BR	$\frac{3}{16}$
BB	$\frac{1}{16}$

Explanation: This problem involves defining a random variable for counting red marbles in two draws with replacement. A random variable X assigns numerical values to the outcomes of an experiment. The possible outcomes when drawing twice are: RR (both red), RB (red then blue), BR (blue then red), and BB (both blue). These outcomes map to X-values: X=2 for RR (two reds), X=1 for RB and BR (one red each), and X=0 for BB (no reds). To find probabilities, we calculate: P(X=0) = P(BB) = (1/4)(1/4) = 1/16, P(X=1) = P(RB) + P(BR) = (3/4)(1/4) + (1/4)(3/4) = 6/16, and P(X=2) = P(RR) = (3/4)(3/4) = 9/16. Option A correctly shows this complete distribution with all X-values and their probabilities summing to 1. A common misconception is confusing the outcome table (option C) with the probability distribution of X.

Question 16

A game uses one draw from a bag containing 2 green tokens and 2 yellow tokens. One token is drawn at random. Let $X$ be your winnings (in dollars): if you draw green, you win $2; if you draw yellow, you lose $1 (so $X=-1 $). Which table correctly represents the probability distribution of$ X$?

$X$ $P(X)$
$-1$ $\frac{1}{2}$
$0$ $\frac{1}{2}$
$X$ $P(X)$
$-1$ $\frac{1}{2}$
$2$ $\frac{1}{2}$
(correct answer)
Outcome Probability
green $\frac{1}{2}$
yellow $\frac{1}{2}$
$X$ $P(X)$
$-1$ $\frac{1}{4}$
$2$ $\frac{3}{4}$

$X$	$P(X)$
$-1$	$\frac{1}{2}$
$0$	$\frac{1}{2}$

$X$	$P(X)$
$-1$	$\frac{1}{2}$
$2$	$\frac{1}{2}$

Explanation: This problem defines X as winnings in dollars based on token color drawn. A random variable assigns numerical values to outcomes - here the token colors map to dollar amounts. The bag has 2 green and 2 yellow tokens (4 total), so each color has probability 1/2. The mapping is: green token gives X=$2 (win

2), and yellow token gives X=-

1 (lose $1). Therefore: P(X=2) = P(green) = 2/4 = 1/2 and P(X=-1) = P(yellow) = 2/4 = 1/2. Option A correctly shows this distribution with X representing actual winnings (including the negative value for a loss) and their probabilities. Note that X takes only two values: 2 and -1, with no other possibilities. Option D incorrectly shows colors instead of winnings, illustrating the difference between outcomes and the random variable's numerical values.

Question 17

A fair six-sided die is rolled once. Let $X$ be the number of points scored, where you score 1 point if the die shows 1, 2, or 3, and you score 2 points if the die shows 4, 5, or 6. Which table correctly represents the probability distribution of $X$ ?

Outcome Probability
1,2,3 $\frac{3}{6}$
4,5,6 $\frac{3}{6}$
$X$ $P(X)$
1 $\frac{1}{6}$
2 $\frac{5}{6}$
$X$ $P(X)$
1 $\frac{1}{2}$
2 $\frac{1}{2}$
(correct answer)
$X$ $P(X)$
1 $\frac{1}{2}$
2 $\frac{1}{2}$
3 $\frac{1}{6}$

Explanation: This problem defines X as points scored based on die roll outcomes. A random variable assigns numerical values to outcomes - here die faces are mapped to point values. The mapping is: faces {1,2,3} each give X=1 point, and faces {4,5,6} each give X=2 points. Since each face has probability 1/6 on a fair die, we calculate: P(X=1) = P(rolling 1,2, or 3) = 3/6 = 1/2 and P(X=2) = P(rolling 4,5, or 6) = 3/6 = 1/2. Option A correctly shows this complete distribution with the two possible point values and their equal probabilities. The distribution is valid because X can only equal 1 or 2 (no other point values are possible) and the probabilities sum to 1. Option D incorrectly groups outcomes without showing the actual X-values, demonstrating the distinction between die faces and point scores.

Question 18

A fair six-sided die is rolled once. Define the random variable $X$ as the number of even results obtained in this one roll (so $X=1$ if the die shows an even number and $X=0$ otherwise). Which table correctly represents the probability distribution of $X$ ?

$X$ $P(X)$
0 $\frac{1}{2}$
1 $\frac{2}{3}$
$X$ $P(X)$
0 $\frac{1}{2}$
1 $\frac{1}{2}$
2 $\frac{1}{6}$
$X$ $P(X)$
0 $\frac{2}{3}$
1 $\frac{1}{3}$
$X$ $P(X)$
0 $\frac{1}{2}$
1 $\frac{1}{2}$
(correct answer)

Explanation: This problem defines X as the number of even results in one die roll, which can only be 0 or 1. A random variable assigns numbers to outcomes - here X indicates whether the die shows an even number. The die outcomes {1,2,3,4,5,6} map to X-values: X=0 for odd outcomes {1,3,5} and X=1 for even outcomes {2,4,6}. Since a fair die has equal probability 1/6 for each face, we calculate: P(X=0) = P(odd) = 3/6 = 1/2 and P(X=1) = P(even) = 3/6 = 1/2. Option A correctly shows this complete distribution with both possible X-values and their equal probabilities. The key insight is that in a single roll, you either get an even number (X=1) or you don't (X=0) - there's no possibility of getting multiple even numbers, so X cannot equal 2.

Question 19

A box contains 1 card labeled 0, 2 cards labeled 1, and 1 card labeled 2. One card is drawn at random. Let $X$ be the number shown on the drawn card. Which table correctly represents the probability distribution of $X$ ?

$X$ $P(X)$
0 $\frac{1}{4}$
1 $\frac{1}{2}$
2 $\frac{1}{2}$
$X$ $P(X)$
0 $\frac{1}{4}$
2 $\frac{1}{4}$
$X$ $P(X)$
0 $\frac{1}{4}$
1 $\frac{1}{4}$
2 $\frac{1}{2}$
$X$ $P(X)$
0 $\frac{1}{4}$
1 $\frac{1}{2}$
2 $\frac{1}{4}$
(correct answer)

Explanation: This problem involves drawing a card where X equals the number shown on the card. A random variable assigns numbers to outcomes - here the card labels directly become the X-values. The box contains 4 cards total: one labeled 0, two labeled 1, and one labeled 2. Since we draw randomly, each card has equal probability of being selected. Therefore: P(X=0) = 1/4 (one card out of four), P(X=1) = 2/4 = 1/2 (two cards out of four), and P(X=2) = 1/4 (one card out of four). Option A correctly displays this complete distribution with all possible X-values and their probabilities summing to 1. The key insight is that when multiple cards show the same number, we combine their probabilities. Option D incorrectly omits X=1, making it an incomplete distribution since cards labeled 1 exist in the box.

Question 20

A fair die is rolled once. The sample space is $\{1,2,3,4,5,6\}$ . Define the random variable $X$ as the number of dots showing if the roll is 1, 2, or 3, and define $X=3$ if the roll is 4, 5, or 6. (So multiple outcomes can map to the same $X$ value.) Which table correctly represents the probability distribution of $X$ ?

| $X$ | $P(X)$ |\n|---|---|\n| $1$ | $\tfrac{1}{6}$ |\n| $2$ | $\tfrac{1}{6}$ |\n| $3$ | $\tfrac{4}{6}$ | (correct answer)
| $X$ | $P(X)$ |\n|---|---|\n| $1$ | $\tfrac{1}{6}$ |\n| $2$ | $\tfrac{1}{6}$ |\n| $3$ | $\tfrac{1}{6}$ |
| $X$ | $P(X)$ |\n|---|---|\n| $1$ | $\tfrac{1}{6}$ |\n| $2$ | $\tfrac{1}{6}$ |\n| $3$ | $\tfrac{3}{6}$ |\n| $4$ | $\tfrac{1}{6}$ |
| $X$ | $P(X)$ |\n|---|---|\n| $1$ | $\tfrac{1}{6}$ |\n| $2$ | $\tfrac{1}{6}$ |\n| $3$ | $\tfrac{1}{2}$ |

Explanation: This problem defines a random variable with a special rule for higher die rolls. A random variable X maps outcomes to numbers: for rolls 1, 2, 3, X equals the number shown; for rolls 4, 5, 6, X equals 3. This gives us X(1) = 1, X(2) = 2, X(3) = 3, X(4) = 3, X(5) = 3, X(6) = 3. Four outcomes (3, 4, 5, 6) map to X = 3, so P(X = 3) = 4/6. The values X = 1 and X = 2 each have one outcome, giving P(X = 1) = 1/6 and P(X = 2) = 1/6. The correct distribution shows these three X-values with probabilities summing to 1. Students might miss that multiple outcomes can map to the same X-value. To solve: carefully apply the mapping rule to each outcome → group by X-value → calculate probabilities.

$X$	$P(X)$
0	$\tfrac{48}{52}\cdot\tfrac{47}{51}$
1	$\tfrac{4}{52}\cdot\tfrac{48}{51}$
2	$\tfrac{4}{52}\cdot\tfrac{3}{51}$