HSS.MD.A.1Common CoreMathStatistics and ProbabilityGrades 9-12
HSS.MD.A.1: Random Variables and Their Probability Distributions
In plain English: HSS.MD.A.1 is an advanced (+) Common Core statistics and probability standard that asks students to define a random variable by giving every outcome in a sample space a number, such as the number of heads in three flips. Students then graph the probability distribution of that variable with the displays used for data, such as a histogram. It is usually taught in a statistics course or Precalculus.
(+) Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space; graph the corresponding probability distribution using the same graphical displays as for data distributions.
Common Core State Standards for Mathematics · Domain: Using Probability to Make Decisions (MD) · Cluster: Calculate expected values and use them to solve problems Also written as HSS-MD.A.1 or S-MD.1 · Official standard
Students learn that a random variable is a rule that gives each outcome of a chance process exactly one number, chosen because that number is the quantity of interest: the number of heads, the points scored on a spin, the larger of two dice. Once outcomes are turned into numbers, outcomes with the same value form an event, and adding their probabilities gives the probability distribution of the variable.
The second half of the lesson treats that distribution the way students already treat data: they draw it as a probability histogram, read probabilities from the bars, and describe its shape as symmetric or skewed. Students compare a histogram of theoretical probabilities with a relative frequency histogram from their own dice rolls, which shows why the same display works for both.
Learning Objectives
By the end of this lesson, students will be able to:
Define a random variable in words for a quantity of interest and give its value for every outcome in a sample space
Group outcomes into events such as X = 2 and find the probability of each value of the variable
Write a probability distribution as a table and check that the probabilities are between 0 and 1 and add to 1
Graph the distribution as a probability histogram and describe its shape as they would describe a data distribution
Prior Knowledge Required
Students should already be comfortable with:
Listing sample spaces with organized lists, tables and tree diagrams 7.SP.C.8
Uniform and non-uniform probability models 7.SP.C.7
Drawing and reading histograms and dot plots of data HSS.ID.A.1
Describing events as subsets of a sample space HSS.CP.A.1
Hand each pair a coin and pose the question below. Give two minutes of quiet work before pairs compare.
Warm-Up Prompt
"Flip a coin three times and write the result as letters, such as HTH. How many different results are possible? Now invent two different numbers you could record about each result."
Build the list of 8 results on the board with the class (2 × 2 × 2 = 8). Collect the numbers students invented: number of heads, number of tails, the length of the longest run, 1 if the first flip is heads and 0 if not. Point out that letters like HTH cannot be added or graphed on a number line, but the numbers can. Each suggestion is a different rule that turns an outcome into a number, and each rule is a different random variable.
Direct Instruction20 minutes
Part 1: Defining a random variable. A random variable, written with a capital letter such as X, assigns one number to each outcome in the sample space. Always define it in words, with units if there are any: "X = the number of heads in three flips." The statement X = 2 then names an event, here {HHT, HTH, THH}. Use Diagram 1 to show the rule as arrows from outcomes to numbers: every outcome has exactly one arrow, but several outcomes can land on the same number. Then give the steps:
List the sample space and decide whether its outcomes are equally likely.
Define the variable in words and give it a capital letter.
Write the value of the variable next to each outcome.
Group the outcomes by value: each group is an event, and its probability is the sum of the probabilities of its outcomes.
Check the distribution: every probability is between 0 and 1, and all of them add to 1.
Graph it as a probability histogram: values on the horizontal axis, bars of equal width centered on each value, and bar heights equal to the probabilities.
Counting equally likely outcomes
Flip a fair coin three times and let X be the number of heads. All 8 outcomes have probability 1/8.
Roll two fair number cubes and let D be the larger number minus the smaller one, with D = 0 for doubles. Count the 36 outcomes for each value (table below).
Equation: D = 0, 1, 2, 3, 4, 5 with probabilities 6/36, 10/36, 8/36, 6/36, 4/36, 2/36
Numbers for outcomes that are not numbers
A spinner is half red, one quarter blue and one quarter green. A player scores R = 0 points on red, 5 points on blue and 10 points on green.
Roll two fair number cubes and let M be the larger of the two numbers (the shared number for doubles). M = m when one cube shows m and the other shows m or less.
Equation: P(M = m) = (2m - 1)/36 for m = 1, 2, ..., 6, so the bars rise from 1/36 to 11/36 (Diagram 2)
Outcomes of two number cubes grouped by the value of D
Part 2: Graphing the distribution. Remind students how they drew a histogram of data: values along the axis and bar heights showing how often each value occurred. A probability histogram is built the same way, with bar heights showing probabilities instead of counts. With bars of width 1, the total area is 1. Sketch the histograms of X and D quickly, then show Diagram 2 for M. Describe each shape in the language used for data: X is symmetric, D is skewed right (its tail stretches toward large differences), and M is skewed left. Stress that D and M come from the same 36 outcomes, yet they have different distributions: the sample space alone does not tell you the graph, the definition of the variable does.
Guided Practice15 minutes
Pairs work through one problem with you, one step at a time: a bag holds 3 red tiles and 2 blue tiles, and two tiles are drawn without replacement. Let X be the number of red tiles drawn. Have students label the tiles R1, R2, R3, B1, B2 and list the 10 equally likely pairs. Then they group the pairs: one pair (B1, B2) gives X = 0, six pairs give X = 1, and three pairs give X = 2, so the distribution is 1/10, 6/10 and 3/10. Pairs draw the probability histogram on grid paper with a labeled vertical scale.
Follow up: define Y = the number of blue tiles drawn. Ask pairs to predict the histogram of Y before computing it (it is the mirror image: 3/10, 6/10, 1/10 at Y = 0, 1, 2). Listen for two errors: giving each of the three values probability 1/3, and mixing ordered and unordered pairs in the same list.
Independent Practice10-15 minutes
Students work alone. A coin is flipped and a number cube is rolled at the same time, giving 12 equally likely outcomes such as (H, 4). Define T = the number on the cube if the coin shows heads, and T = 0 if the coin shows tails. Students list the outcomes for each value of T, write the distribution (P(T = 0) = 6/12 = 1/2 and P(T = k) = 1/12 for k = 1, ..., 6) and draw its histogram. Then each student defines a second random variable of their own on the same 12 outcomes, graphs it, and gives only the graph to a partner, who must suggest a rule that produces it.
Closure5 minutes
Exit ticket: (1) Roll one fair number cube and let Y = 1 if the number is a multiple of 3 and Y = 0 otherwise. Give the distribution of Y and sketch its histogram. (Answer: P(Y = 1) = 2/6 = 1/3 and P(Y = 0) = 2/3.) (2) In one sentence, explain why the arrows in Diagram 1 make the random variable a function.
Differentiation Strategies
For Struggling Students
Give a printed 6-by-6 grid of dice outcomes and have students write the value of the variable in each cell before counting, then shade cells with equal values in the same color
Provide a four-column template: value, outcomes, count, probability, with the total row already drawn
Start with one coin flip and one number cube before moving to three flips or two cubes
For Advanced Students
Define N = the number of flips of a coin until the first head. Graph the first six bars of P(N = k) = (1/2)k and explain why the bars still add to 1 even though N has no largest value
Find two different random variables on the two-cube sample space that have exactly the same distribution, and explain why
Explain why the time until a bus arrives cannot be graphed with bars at single values, and what kind of graph might replace them
Assessment Guidance
What to Look For
Listen for a definition in words before any numbers: "X is the number of red tiles drawn" rather than "X is red." Check that every outcome gets exactly one value and that no outcome is left out; a quick test is that the counts add to the size of the sample space. On graphs, look for a labeled probability scale, bars of equal width centered on the values, and heights that match the table. When students describe shape, ask them to use the same words they use for data: symmetric, skewed left, skewed right, and the location of the peak.
02
Classroom Activities
3 Activities
1
One Sample Space, Four Random Variables
20 minGroups of 3-4
Every group uses the same chance process, two spins of a spinner with four equal sections numbered 1 to 4, but each group defines a different random variable on it. Comparing the four histograms shows that the variable, not the sample space, decides the shape of the distribution.
Variable Cards
Card E: E = the number of spins that land on an even number
Card F: F = the first spin minus the second spin
Card G: G = 1 if the two spins match and G = 0 if they do not
Card H: H = the product of the two spins
Procedure
Each group draws a 4-by-4 grid of the 16 equally likely outcomes and writes the value of its variable in every cell
The group builds a table of values, counts and probabilities, and checks that the counts add to 16
The group draws the probability histogram on grid paper and writes one sentence about its shape
Gallery walk: groups visit the other posters and write one difference they notice on a sticky note
F: values -3 to 3 with counts 1, 2, 3, 4, 3, 2, 1 out of 16 (symmetric, peak at 0)
G: P(G = 1) = 4/16 and P(G = 0) = 12/16
H: values 1, 2, 3, 4, 6, 8, 9, 12, 16 with counts 1, 2, 2, 3, 2, 2, 1, 2, 1 (gaps between values, bars at the listed values only)
Modification for Distance Learning
Share the 4-by-4 grid as a spreadsheet. Each breakout room fills the cells with its variable, uses a count formula for each value, and inserts a column chart of the probabilities.
2
Theory Versus Data: Dice Differences
20 minPairs
Pairs collect real data for the random variable D (larger number minus smaller) from worked example 2, draw a relative frequency histogram, and lay it next to the probability histogram. This shows that the same kind of display serves both a data distribution and a probability distribution.
Procedure
Each pair rolls two number cubes 50 times and records D for every roll in a tally table
The pair divides each tally by 50 and draws a relative frequency histogram on grid paper
On the same axes, in a second color, the pair draws the probability histogram from the table in the lesson (6/36, 10/36, 8/36, 6/36, 4/36, 2/36)
The class pools all tallies on the board, and one student computes the pooled relative frequencies
Discussion Questions
Which bars of your data histogram are farthest from the probability histogram? Is the cube unfair, or is this chance variation?
Is the pooled class histogram closer to the probability histogram than your pair's histogram? Why would that happen?
What is the same about the two graphs, and what does the height of a bar mean in each?
Challenge Variation
Use a spreadsheet random number generator to simulate 1,000 rolls of two cubes and compare the new relative frequency histogram with the class data.
3
Match the Histogram
15 minPairs
Pairs receive 8 cards: 4 describe a chance process and a random variable, and 4 show a probability histogram. Pairs match each variable to its histogram and justify the match by finding at least one probability from the sample space.
Variable Cards
(a) The number showing on one roll of a fair number cube
(b) The number of sixes in two rolls of a fair number cube
(c) The number of coin flips up to and including the first head, where you stop after 4 flips no matter what
(d) The total of two spins of a spinner with three equal sections numbered 1, 2 and 3
Histogram Cards
I: bars at 1, 2, 3, 4 with heights 1/2, 1/4, 1/8, 1/8
II: bars at 2, 3, 4, 5, 6 with heights 1/9, 2/9, 3/9, 2/9, 1/9
III: six bars of equal height 1/6 at 1 through 6
IV: bars at 0, 1, 2 with heights 25/36, 10/36, 1/36
Print each histogram card as a drawn graph with a labeled probability scale; the heights are listed here for the teacher. Key: (a) III, (b) IV, (c) I, (d) II.
Discussion Questions
Why is the last bar of card I the same height as the bar before it?
Which two histograms are symmetric? What about the chance process makes them symmetric?
Challenge Variation
Each pair writes a new variable card and draws its histogram on a separate card, then trades both cards, shuffled, with another pair.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: A Random Variable Assigns a Number to Each Outcome
The random variable X = the number of heads sends each of the 8 outcomes of three coin flips to exactly one number. Outcomes that land on the same number form an event, so P(X = 2) = 3/8 because three outcomes land on 2.
Diagram 2: Probability Histogram of the Larger of Two Dice
Each bar is centered on a value of M and its height is P(M = m) = (2m - 1)/36, drawn to scale on the vertical axis. Like a data histogram, it has a shape: the tall bars are on the right and the tail is on the left, so the distribution is skewed left.
04
Homework Assignment
~30 min
HSS.MD.A.1 Homework: Random Variables and Probability Histograms
Directions: For every problem, define the random variable in words before you calculate. Show the sample space or explain how you counted it. Write each distribution as a table, check that the probabilities add to 1, and draw histograms on grid paper with labeled axes.
Part 1: Defining Random Variables (Problems 1-3)
A fair coin is tossed four times, and X is the number of tails. (a) How many outcomes are in the sample space? (b) List the outcomes for X = 1 and for X = 2. (c) Write the probability distribution of X as a table.
A club with 4 juniors and 2 seniors chooses 2 of its members at random to attend a meeting. Let Y be the number of seniors chosen. List the 15 possible pairs of members, then find the probability distribution of Y.
In a board game you roll one fair number cube. A roll of 1 or 2 moves you 0 spaces, a roll of 3, 4 or 5 moves you 2 spaces, and a roll of 6 moves you 5 spaces. Define a random variable S for this game, give its probability distribution, and draw its probability histogram.
Part 2: Graphing Distributions (Problems 4-6)
Two fair number cubes are rolled. Let G be the number of cubes that show a 5 or a 6. Find the probability distribution of G, draw its probability histogram, and describe its shape.
The probability histogram of a random variable W has bars at W = 0, 1, 2, 3 and 4 with heights 0.10, 0.25, h, 0.20 and 0.05. (a) Find h. (b) Find P(W ≥ 3). (c) Which value of W is most likely? (d) Find the probability that W is an odd number.
A student says that for two flips of a fair coin, the number of heads H has P(H = 0) = P(H = 1) = P(H = 2) = 1/3, because H has three possible values. Use the sample space to explain the error, give the correct distribution, and sketch both histograms so their shapes can be compared.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Sample Space and Definition
Variable defined in words; sample space complete
Definition vague or one outcome missing
No definition or sample space
Distribution
All probabilities correct and adding to 1
One or two probabilities wrong
Most probabilities wrong
Histogram
Labeled axes, equal widths, heights to scale
Correct heights but missing labels or scale
Missing or does not match the table
Explanation
Shape and errors explained clearly
Explanation partly correct
No explanation
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which of these is a random variable for the chance process of flipping a coin five times?
Answer: B
A random variable gives a number to every outcome, and the number of heads does that: HHTHT gets 3, TTTTT gets 0, and so on. Choice A is a single outcome, not a rule. Choice C is a fixed probability (1/2), not a quantity that changes from outcome to outcome. Choice D is the sample space itself.
Question 2 of 20 · Multiple Choice
A fair number cube is rolled twice. Let X be the number of rolls that show an odd number. How many of the 36 equally likely outcomes give X = 1?
Answer: C
X = 1 happens for odd then even (3 × 3 = 9 outcomes) or even then odd (another 9), so 18 outcomes. Choice A counts only one order. Choice D counts every outcome with at least one odd roll, which also includes the 9 outcomes where both rolls are odd (X = 2).
Question 3 of 20 · Multiple Choice
A fair coin is flipped 5 times, and X is the number of heads. What is P(X = 0)?
Answer: D
There are 25 = 32 equally likely outcomes, and only TTTTT has 0 heads, so P(X = 0) = 1/32. Choice B treats the six possible values 0 through 5 as equally likely, but they are not: many more outcomes have 2 or 3 heads than 0.
Question 4 of 20 · Multiple Choice
Which table is a valid probability distribution?
Answer: D
In D every probability is between 0 and 1 and they add to 1. Choice A adds to 1.1. Choice B adds to 1 but contains a negative probability, which is impossible. Choice C adds to only 0.6.
Question 5 of 20 · Multiple Choice
A random variable X takes the values 1, 2, 3 and 4 with probabilities 0.15, 0.35, k and 0.30. What is k?
Answer: A
The probabilities must add to 1: 0.15 + 0.35 + k + 0.30 = 1, so k = 1 - 0.80 = 0.20. Choice D is the sum of the known probabilities, not what is left over. Choice B assumes the four values share the probability equally.
Question 6 of 20 · Multiple Choice
In a probability histogram for a random variable, what does the height of each bar show?
Answer: B
The horizontal position of a bar is the value, and its height is the probability of that value. Choice C mixes up the two axes. Choice D describes a frequency histogram of collected data, which looks similar but shows what happened in trials rather than theoretical probabilities.
Question 7 of 20 · Multiple Choice
A random variable has P(X = 0) = 0.4, P(X = 1) = 0.3, P(X = 2) = 0.2 and P(X = 3) = 0.1. How would you describe the shape of its probability histogram?
Answer: D
The tallest bar is at 0 and the bars get shorter to the right, so the tail stretches to the right: skewed right. Choice B names the direction of the tall bars instead of the tail. Skew is named for the side with the tail, just as for data.
Question 8 of 20 · Multiple Choice
A probability histogram has bars at X = 1, 2, 3, 4 and 5 with heights 0.1, 0.2, 0.4, 0.2 and 0.1. What is P(X ≥ 4)?
Answer: B
Add the heights of the bars at 4 and 5: 0.2 + 0.1 = 0.3. Choice A reads only the bar at X = 4. Choice C is P(X ≥ 3), which includes the bar at 3 as well.
Question 9 of 20 · Multiple Choice
A spinner lands on blue with probability 1/2, yellow with probability 1/3 and purple with probability 1/6. A player scores 1 point for blue, 3 points for yellow and 6 points for purple. Let S be the score. Which is the probability distribution of S?
Answer: A
Each score comes from exactly one color, so each value of S takes that color's probability. Choice B ignores the different sizes of the sections. Choice D makes the probabilities proportional to the points, confusing the value of the variable with its probability.
Question 10 of 20 · Multiple Choice
A bag holds 4 green marbles and 1 yellow marble. Two marbles are drawn without replacement, and Y is the number of yellow marbles drawn. What is P(Y = 1)?
Answer: B
There are 10 equally likely pairs of marbles, and 4 of them contain the yellow marble, so P(Y = 1) = 4/10 = 2/5. Choice A is the chance of yellow on a single draw. Choice D is the answer if the first marble were put back before the second draw.
Question 11 of 20 · Multiple Choice
One card is drawn from a standard 52-card deck. Let X = 1 if the card is a heart and X = 0 otherwise. What is P(X = 0)?
Answer: C
39 of the 52 cards are not hearts, so P(X = 0) = 39/52 = 3/4. Choice A is P(X = 1). Choice B assumes the two values of X are equally likely. Choice D confuses suits with ranks.
Question 12 of 20 · Multiple Choice
A class rolls a fair number cube 60 times and graphs the fraction of rolls that showed each number. How does this graph compare with the probability histogram for one roll?
Answer: A
The class graph is a relative frequency histogram of data: same display, but the heights come from what happened in 60 rolls, so some bars will be above 1/6 and some below. Choice C would be true for counts, not for fractions of the 60 rolls, which always add to 1.
Question 13 of 20 · Multiple Choice
A fair four-sided die numbered 1 to 4 is rolled twice, and S is the sum of the two rolls. What is P(S = 5)?
Answer: D
Of the 16 equally likely ordered outcomes, (1,4), (4,1), (2,3) and (3,2) give S = 5, so P(S = 5) = 4/16 = 1/4. Choice A treats the 7 possible sums (2 through 8) as equally likely. Choice B counts only the two unordered pairs {1,4} and {2,3}.
Question 14 of 20 · Multiple Choice
A coin is flipped three times, and Y = (number of heads) - (number of tails). Which of these is not a possible value of Y?
Answer: C
The number of heads h and tails 3 - h give Y = 2h - 3, which is -3, -1, 1 or 3. Y is always odd, so 0 is impossible. Choices A and D come from TTT and HHH, and B from one head and two tails.
Question 15 of 20 · Short Answer
Two fair number cubes are rolled, and X is the sum of the two numbers. List the outcomes that give X = 4 and find P(X = 4).
The outcomes are (1,3), (2,2) and (3,1), so P(X = 4) = 3/36 = 1/12. A common error is to count (1,3) and (3,1) as one outcome; the two cubes are different, so they are two of the 36 outcomes.
Question 16 of 20 · Short Answer
A bag has 2 red chips and 3 white chips. You draw a chip, put it back, and draw again. Let R be the number of red chips drawn. Find the probability distribution of R and describe the shape of its histogram.
P(red) = 2/5 on each draw. P(R = 0) = (3/5)² = 9/25, P(R = 1) = 2(2/5)(3/5) = 12/25, P(R = 2) = (2/5)² = 4/25. The check 9 + 12 + 4 = 25 works. The tallest bar is at R = 1 and the bar at 2 is the shortest, so the histogram is slightly skewed right.
Question 17 of 20 · Short Answer
A random variable X has the values 2, 4, 6 and 8 with probabilities 0.1, 0.3, 0.4 and 0.2. Describe its probability histogram, then find P(X > 4).
Four bars at 2, 4, 6 and 8 with heights 0.1, 0.3, 0.4 and 0.2; the tallest bar is at 6, so 6 is the most likely value. P(X > 4) = 0.4 + 0.2 = 0.6. The bar at 4 is not included because the inequality is strict.
Question 18 of 20 · Short Answer
Explain why this table is not a probability distribution: x = 0, 1, 2, 3 with P = 0.2, 0.5, 0.4, -0.1.
The probabilities do add to 1 (0.2 + 0.5 + 0.4 - 0.1 = 1), but P(X = 3) = -0.1 is negative, and no probability can be less than 0. A histogram would need a bar of negative height, which has no meaning.
Question 19 of 20 · Short Answer
A fair coin is flipped three times. Let Y = 1 if all three flips land the same way and Y = 0 otherwise. Give the probability distribution of Y.
Only HHH and TTT make all three flips match, so P(Y = 1) = 2/8 = 1/4 and P(Y = 0) = 6/8 = 3/4. A variable like Y that only takes the values 0 and 1 is called an indicator variable.
Question 20 of 20 · Short Answer
A computer simulates 500 rolls of a fair number cube. In the relative frequency histogram, the bar for 6 has height 0.182. Does this show that the cube is unfair? What should happen to the histogram if the computer simulates 50,000 rolls?
No. The probability histogram has every bar at 1/6 ≈ 0.167. A height of 0.182 means 91 sixes instead of about 83, a difference that chance alone can easily produce in 500 rolls. With 50,000 rolls, the relative frequencies should settle much closer to 1/6, so the data histogram should look almost flat, like the probability histogram.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSS.MD.A.1 mean?
HSS.MD.A.1 means students can turn the outcomes of a chance process into numbers and graph the result. They define a random variable, such as the number of defective parts in a sample, by giving every outcome in the sample space a value. Then they find the probability of each value and draw the probability distribution as a histogram, the same way they would graph data.
What is a random variable in simple terms?
A random variable is a rule that gives one number to each possible outcome of a chance process. The outcome itself might be a list of letters, a color or a card, but the random variable records a number about it: how many heads, how many points, how many minutes. It is called random because you do not know its value until the chance process happens.
Is HSS.MD.A.1 taught in Algebra 2 or Statistics?
It varies by school. HSS.MD.A.1 is marked (+), which Common Core uses for additional mathematics that students should learn to take advanced courses. It is usually taught in a statistics course or Precalculus, and some Algebra II courses include it before expected value.
What is the difference between a probability distribution and a data distribution?
A data distribution shows what actually happened in a set of observations, such as 40 rolls of a cube. A probability distribution shows what the chance model predicts in the long run, before any rolls. Both list values and how common each value is, so both can be graphed with a histogram. As the number of trials grows, the relative frequency histogram of the data tends to get closer to the probability histogram.
How do you graph a probability distribution?
Use a probability histogram. Put the values of the random variable on the horizontal axis and probability on the vertical axis. Draw a bar of equal width centered on each value, with height equal to that value's probability. Label both axes and use a scale that starts at 0. For a variable with only a few values, a bar graph with gaps between the bars is also common.
Why is a random variable called a function?
Because it matches every outcome in the sample space with exactly one number, which is the definition of a function. The sample space is the domain and the possible values are the range. Several outcomes can map to the same number, which is why the probability of a value is often the sum of several outcome probabilities.
What mistakes do students make with random variables?
A common one is assuming that every value of the variable is equally likely just because the outcomes are. Another is defining the variable with a word instead of a number ("X = heads"). Students also forget that the two cubes in a roll are different, so (2, 5) and (5, 2) are two outcomes, or they draw histograms whose bar heights do not add to 1.
Do the probabilities in a probability distribution always add to 1?
Yes. Every outcome in the sample space is assigned to exactly one value of the random variable, so the events X = x for the different values cover the whole sample space without overlap. Their probabilities must therefore add to 1, and each one must be between 0 and 1. Checking the sum is a quick way to catch a missing or double-counted outcome.
What is the difference between a discrete and a continuous random variable?
A discrete random variable has values you can list, such as 0, 1, 2 or 3, and its distribution can be shown with a bar for each value. A continuous random variable, such as a waiting time, can take any value in an interval, so probabilities come from areas under a curve instead of bar heights. HSS.MD.A.1 works with discrete random variables; normal curves appear in HSS.ID.A.4.
How does HSS.MD.A.1 connect to expected value?
Directly: HSS.MD.A.2, the next standard in the same cluster, asks students to calculate the expected value of a random variable and interpret it as the mean of its probability distribution. The distribution built in HSS.MD.A.1 is exactly the input for that calculation. HSS.MD.A.3 and HSS.MD.A.4 then have students build distributions from theoretical and from empirical probabilities.
07
Related Standards
5 standards
These standards connect to HSS.MD.A.1: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.SP.C.8Prerequisite
Find probabilities of compound events with organized lists, tables, tree diagrams
