Defining Random Variables and Probability Distributions
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Statistics › Defining Random Variables and Probability Distributions
A game is played by rolling a fair die once. You win $5$ if the roll is 1 or 2, win $1$ if the roll is 3, 4, or 5, and lose $2$ if the roll is 6. Let $X$ be your winnings (in dollars). Which table correctly represents the probability distribution of $X$?
| $X$ | $P(X)$ |
|---|---|
| 5 | $\tfrac{2}{6}$ |
| 1 | $\tfrac{3}{6}$ |
| $X$ | $P(X)$ |
|---|---|
| 5 | $\tfrac{2}{6}$ |
| 1 | $\tfrac{3}{6}$ |
| -2 | $\tfrac{1}{6}$ |
| $X$ | $P(X)$ |
|---|---|
| 1 | $\tfrac{1}{6}$ |
| 2 | $\tfrac{1}{6}$ |
| 3 | $\tfrac{1}{6}$ |
| 4 | $\tfrac{1}{6}$ |
| 5 | $\tfrac{1}{6}$ |
| 6 | $\tfrac{1}{6}$ |
| $X$ | $P(X)$ |
|---|---|
| 5 | $\tfrac{1}{6}$ |
| 1 | $\tfrac{3}{6}$ |
| -2 | $\tfrac{2}{6}$ |
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. For this die game, $X$ is winnings: rolls 1-2 map to 5, 3-5 to 1, 6 to -2. Thus, outcomes group into these $X$ values. Probabilities: $P(X=5) = \frac{2}{6}$, $P(X=1) = \frac{3}{6}$, $P(X=-2) = \frac{1}{6}$, from equal face probabilities. It's complete and valid, listing all $X$ (5,1,-2) summing to 1. Misconception: confusing uniform die outcomes with uniform $X$; $X$ probabilities vary by grouping. Method: list outcomes, assign $X$, sum probabilities per $X$.
A fair coin is flipped three times. Define the random variable $X$ as the number of tails obtained. Which table correctly represents the probability distribution of $X$?
| $X$ | $P(X)$ |
|---|---|
| 0 | $\tfrac{1}{8}$ |
| 1 | $\tfrac{3}{8}$ |
| 2 | $\tfrac{3}{8}$ |
| 3 | $\tfrac{1}{8}$ |
| $X$ | $P(X)$ |
|---|---|
| 0 | $\tfrac{1}{8}$ |
| 1 | $\tfrac{3}{8}$ |
| 2 | $\tfrac{3}{8}$ |
| $X$ | $P(X)$ |
|---|---|
| 0 | $\tfrac{1}{8}$ |
| 1 | $\tfrac{3}{8}$ |
| 2 | $\tfrac{1}{8}$ |
| 3 | $\tfrac{3}{8}$ |
| $X$ | $P(X)$ |
|---|---|
| 0 | $\tfrac{1}{4}$ |
| 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 three coin flips, X counts tails, so each sequence maps to X=0 to 3 by tail count. Outcomes like TTH, THT, HTT all map to X=2. Probabilities: P(X=0)=1/8, P(X=1)=3/8, P(X=2)=3/8, P(X=3)=1/8, from 8 equal outcomes. It's complete and valid, covering X=0-3 summing to 1. Misconception: listing all sequences as the distribution; group by X instead. Steps: list flip sequences, count tails for X, combine probabilities.
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 | $\tfrac{1}{4}$ |
| 2 | $\tfrac{1}{4}$ |
| $X$ | $P(X)$ |
|---|---|
| 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 | $\tfrac{1}{4}$ |
| 1 | $\tfrac{1}{2}$ |
| 2 | $\tfrac{1}{2}$ |
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.
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 $X$?
| $X$ | $P(X)$ |
|---|---|
| 1 | $\tfrac{2}{4}$ |
| 2 | $\tfrac{1}{4}$ |
| 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 | $\tfrac{2}{4}$ |
| 3 | $\tfrac{1}{4}$ |
| $X$ | $P(X)$ |
|---|---|
| 0 | $\tfrac{1}{4}$ |
| 1 | $\tfrac{2}{4}$ |
| 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. 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.
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 | $\tfrac{1}{3}$ |
| 0 | $\tfrac{1}{3}$ |
| 2 | $\tfrac{1}{3}$ |
| $X$ | $P(X)$ |
|---|---|
| -1 | $\tfrac{1}{2}$ |
| 0 | $\tfrac{1}{3}$ |
| 2 | $\tfrac{1}{3}$ |
| $X$ | $P(X)$ |
|---|---|
| -1 | $\tfrac{1}{3}$ |
| 0 | $\tfrac{1}{3}$ |
| 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.
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 $X$?
| $X$ | $P(X)$ |
|---|---|
| 0 | $\tfrac{1}{6}$ |
| 1 | $\tfrac{5}{6}$ |
| $X$ | $P(X)$ |
|---|---|
| 0 | $\tfrac{3}{6}$ |
| 1 | $\tfrac{2}{6}$ |
| $X$ | $P(X)$ |
|---|---|
| 0 | $\tfrac{1}{2}$ |
| 1 | $\tfrac{1}{2}$ |
| $X$ | $P(X)$ |
|---|---|
| 2 | $\tfrac{1}{2}$ |
| 4 | $\tfrac{1}{2}$ |
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.
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 | $\tfrac{2}{3}$ |
| $X$ | $P(X)$ |
|---|---|
| R | $\tfrac{2}{3}$ |
| B | $\tfrac{1}{3}$ |
| $X$ | $P(X)$ |
|---|---|
| 0 | $\tfrac{1}{3}$ |
| 1 | $\tfrac{2}{3}$ |
| 2 | $\tfrac{1}{3}$ |
| $X$ | $P(X)$ |
|---|---|
| 0 | $\tfrac{2}{3}$ |
| 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. 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.
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 | $\tfrac{4}{52}\cdot\tfrac{48}{51}+\tfrac{48}{52}\cdot\tfrac{4}{51}$ |
| 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 | $\tfrac{48}{52}$ |
| 1 | $\tfrac{4}{52}$ |
| 2 | $\tfrac{4}{52}\cdot\tfrac{3}{51}$ |
| $X$ | $P(X)$ |
|---|---|
| 0 | $\tfrac{48}{52}\cdot\tfrac{47}{51}$ |
| 1 | $\tfrac{4}{52}\cdot\tfrac{48}{51}$ |
| 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.
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 $X$?
| $X$ | $P(X)$ |
|---|---|
| 2 | $\tfrac{1}{2}$ |
| 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 | $\tfrac{1}{4}$ |
| 5 | $\tfrac{1}{4}$ |
| 8 | $\tfrac{1}{4}$ |
| $X$ | $P(X)$ |
|---|---|
| 2 | $\tfrac{2}{5}$ |
| 5 | $\tfrac{1}{5}$ |
| 8 | $\tfrac{2}{5}$ |
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.
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 | $\tfrac{2}{3}$ |
| $X$ | $P(X)$ |
|---|---|
| $4$ | $\tfrac{1}{3}$ |
| $10$ | $\tfrac{2}{3}$ |
| $X$ | $P(X)$ |
|---|---|
| $10$ | $\tfrac{1}{3}$ |
| $X$ | $P(X)$ |
|---|---|
| $4$ | $\tfrac{2}{3}$ |
| $10$ | $\tfrac{2}{3}$ |
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}$, $P$(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).