HSS.MD.A.3Common CoreMathStatistics and ProbabilityGrades 9-12
HSS.MD.A.3: Theoretical Probability Distributions and Expected Value
In plain English: HSS.MD.A.3 is an advanced (+) Common Core statistics and probability standard that asks students to build the probability distribution of a random variable from theoretical probabilities, found by counting outcomes or using a tree diagram, and then find its expected value. A classic case is the number of correct guesses on a multiple-choice test. It is usually taught in a statistics course or Precalculus.
(+) Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value. For example, find the theoretical probability distribution for the number of correct answers obtained by guessing on all five questions of a multiple-choice test where each question has four choices, and find the expected grade under various grading schemes.
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.3 or S-MD.3 · Official standard
Students develop a probability distribution from a model rather than from data. When a chance process has equally likely outcomes, or independent stages whose probabilities are known, the probability of each value of a random variable can be calculated exactly: by counting outcomes, by multiplying along the branches of a tree diagram, or by counting paths with combinations. Students then use the finished distribution to find the expected value.
The lesson is built around the official example of the standard. Students guess on every question of a five-question multiple-choice test with four choices per question, find the theoretical distribution of the number of correct answers, and compare the expected grade under several grading schemes, including schemes that take off points for wrong answers. The same method is then applied to dice, cards and drawings without replacement.
Learning Objectives
By the end of this lesson, students will be able to:
Identify when a chance process has probabilities that can be calculated theoretically, from equally likely outcomes or independent stages
Develop the probability distribution of a random variable by counting outcomes, using a tree diagram, or counting paths with combinations
Find the theoretical distribution of the number of correct answers when guessing on a multiple-choice test
Find the expected value of the variable, and of a score or grade computed from it, and interpret the result
Prior Knowledge Required
Students should already be comfortable with:
Random variables and probability distributions HSS.MD.A.1
Expected value as the mean of a distribution HSS.MD.A.2
Multiplying probabilities of independent events HSS.CP.A.2
Tree diagrams and organized lists for compound events 7.SP.C.8
Combinations, for the faster counting method HSS.CP.B.9
Tell students that they will take a two-question quiz on a topic nobody in the room has studied, so everyone must guess. Each question has four choices, and exactly one is correct.
Warm-Up Prompt
"You guess on both questions. What is the probability that you get both right? Neither right? Exactly one right? Which is most likely?"
Let pairs argue, then draw the tree in Diagram 2 together. Both right: (1/4)(1/4) = 1/16. Neither right: (3/4)(3/4) = 9/16. Exactly one right happens along two branches, CW and WC, each with probability 3/16, so the total is 6/16. The three probabilities add to 16/16. Many students expect "exactly one" to be the most likely; the tree shows that getting neither right is more likely. Tell students the lesson will scale this up to the five-question test in the standard.
Direct Instruction20-25 minutes
Part 1: When are probabilities theoretical? A theoretical probability comes from a model of how the chance process works: fair dice, a well-shuffled deck, a random guess among four choices, or independent stages. No data are needed. (HSS.MD.A.4 handles the other case, where probabilities are estimated from data.) Present the method:
State the model: which outcomes are equally likely, or which stages are independent and with what probabilities.
Define the random variable and list all of its possible values.
Find each probability: count equally likely outcomes, or multiply along each branch of a tree and add the branches that give the same value. When every branch with k successes has the same probability, multiply that probability by the number of such branches.
Check that the probabilities add to 1.
Find E(X) = Σ x · P(X = x), and the expected value of any score computed from X, by weighting each score by the probability of its value of X.
Equally likely outcomes
Roll two fair number cubes and let S be the sum. Each of the 36 ordered outcomes has probability 1/36, and the number of outcomes with sum s rises from 1 (sum 2) to 6 (sum 7) and falls back to 1 (sum 12).
Equation: P(S = s) = (6 - |s - 7|)/36 for s = 2, ..., 12, and E(S) = 252/36 = 7
Official example: the distribution
A student guesses on all five questions of a multiple-choice test, each with four choices. X is the number of correct answers. Each question is right with probability 1/4, independently, and there are C(5, k) branches with k correct answers.
Equation: P(X = k) = C(5, k)(1/4)k(3/4)5-k, which gives 243, 405, 270, 90, 15 and 1 out of 1024 for k = 0 to 5, and E(X) = 1280/1024 = 1.25
Official example: grading schemes
Use the distribution of X to find the expected grade. The number of wrong answers is 5 - X, so its expected value is 5 - 1.25 = 3.75.
Equation: 20 points per correct answer: E = 20(1.25) = 25 out of 100. One point per correct answer minus 1/4 point per wrong answer: E = 1.25 - 3.75/4 = 0.3125. Minus 1/3 point per wrong answer: E = 1.25 - 3.75/3 = 0
Drawing without replacement
A box holds 6 batteries, 2 of them dead. You take 3 at random, and D is the number of dead batteries. Each of the C(6, 3) = 20 groups of 3 is equally likely.
Theoretical distribution of the number of correct guesses on five 4-choice questions
k correct
Branches C(5, k)
Probability of one branch
P(X = k)
k · P(X = k)
0
1
243/1024
243/1024 ≈ 0.237
0
1
5
81/1024
405/1024 ≈ 0.396
405/1024
2
10
27/1024
270/1024 ≈ 0.264
540/1024
3
10
9/1024
90/1024 ≈ 0.088
270/1024
4
5
3/1024
15/1024 ≈ 0.015
60/1024
5
1
1/1024
1/1024 ≈ 0.001
5/1024
Total
32
1024/1024 = 1
1280/1024 = 1.25
Part 2: Reading the result. Show Diagram 1. The most likely result of guessing is 1 correct answer, the expected value is 1.25, and the histogram is skewed right. Point out that 1.25 = 5 × 1/4: each question adds 1/4 of a correct answer on average, so the expected number correct from guessing on n questions with c choices is n/c. For the grading schemes, stress the idea behind the third one: with a penalty of 1/3 point per wrong answer on a four-choice test, a student who guesses on everything has an expected score of 0, so blind guessing neither helps nor hurts on average. Ask the class which of the three schemes they would choose if they had not studied, and why.
Guided Practice15 minutes
Pairs work one problem step by step with you. A spinner has three equal sections, one red and two blue, and is spun three times. Let B be the number of blue results. Draw the first two levels of the tree together and let pairs finish it. Each branch with k blues has probability (2/3)k(1/3)3-k, and there are 1, 3, 3 and 1 branches for k = 0, 1, 2, 3. So P(B = 0) = 1/27, P(B = 1) = 6/27, P(B = 2) = 12/27 and P(B = 3) = 8/27. Pairs check the sum, then find E(B) = (6 + 24 + 24)/27 = 54/27 = 2 blue results, and confirm it equals 3 × 2/3.
Then give a scoring rule: a player earns 3 points for each blue and loses 2 points for each red. Pairs list the score for each value of B (-6, -1, 4, 9) and find the expected score: (-6 · 1 - 1 · 6 + 4 · 12 + 9 · 8)/27 = 108/27 = 4 points. Watch for pairs who forget that every branch with the same number of blues has the same probability, and for pairs who add probabilities along a branch instead of multiplying.
Independent Practice10-15 minutes
Students work alone on two problems and compare with a partner at the end. (1) A number cube is rolled 3 times, and N is the number of sixes. Develop the distribution (125/216, 75/216, 15/216, 1/216) and find E(N) = 108/216 = 0.5. (2) Four cards numbered 1, 2, 3 and 4 are shuffled, and two are drawn without replacement. T is the sum of the two cards. List the 6 equally likely pairs, develop the distribution (T = 3, 4, 5, 6, 7 with probabilities 1/6, 1/6, 2/6, 1/6, 1/6), and find E(T) = 30/6 = 5.
Closure5 minutes
Exit ticket: A student guesses on a three-question true-false quiz. (1) Develop the distribution of the number of correct answers. (Answer: 1/8, 3/8, 3/8, 1/8 for 0 to 3 correct.) (2) The teacher gives 2 points per correct answer and takes away 1 point per wrong answer. What is the guesser's expected score? (Scores -3, 0, 3, 6, so E = (-3 + 0 + 9 + 6)/8 = 12/8 = 1.5 points.)
Differentiation Strategies
For Struggling Students
Start with the two-question tree and extend it to three questions by drawing every branch, before introducing the count of branches C(n, k)
Give a table template with columns for value, number of branches, probability of one branch, probability and value times probability
Use fractions with the same denominator (all over 1024, all over 27) until the final step, so the sum check is easy
For Advanced Students
Find the penalty per wrong answer that makes the expected score of a guesser 0 on a test with c choices per question, and explain the result
Suppose the student can rule out one wrong choice on every question of the five-question test. Develop the new distribution and compare the expected grades under the three schemes
Show that for any number of guessed questions n, the probabilities C(n, k)(1/4)k(3/4)n-k add to 1, using the binomial theorem
Assessment Guidance
What to Look For
Ask students to state the model before calculating: "each question is a random guess among 4 choices, independent of the others." Check that the probability of each value counts every branch that produces it, not only one; a distribution that does not add to 1 usually means branches were missed. For grading schemes, look for the score written as a function of X (for example X - (5 - X)/4) and the expected value computed from the distribution or from E(X). In interpretations, students should say that 1.25 is an average over many guessers or many tests, not a score that a single student gets.
02
Classroom Activities
3 Activities
1
The Guessing Test
25 minWhole class, then groups of 3-4
Students work through the official example of the standard. They first simulate guessing on a five-question, four-choice test, then build the theoretical distribution of the number of correct answers and use it to compare expected grades under different grading schemes.
Part A: Simulate
Write an answer key of five letters A to D on the board and cover it
Each student "answers" the five questions by rolling a four-sided die (1 = A, 2 = B, 3 = C, 4 = D) or spinning a four-section spinner
Uncover the key; each student counts their correct answers, and the class builds a dot plot of the results
Part B: Build the Theoretical Distribution
Groups explain why every answer is correct with probability 1/4, independently of the others
Groups find how many of the 1024 equally likely answer strings give k correct answers (C(5, k) · 35-k) and complete the distribution
Groups draw the probability histogram next to the class dot plot and compare the shapes
Part C: Grading Schemes
Scheme 1: 20 points per correct answer (a grade out of 100). Expected grade 25
Scheme 2: 1 point per correct answer, minus 1/4 point per wrong answer. Expected score 0.3125
Scheme 3: 1 point per correct answer, minus 1/3 point per wrong answer. Expected score 0
Scheme 4: the test is passed only with 4 or 5 correct answers. P(pass) = 16/1024 = 1/64
For schemes 1 to 3, groups compute the score for each value of X and weight it by P(X = k), then check with E(X) = 1.25
Discussion Questions
Why is the class dot plot not exactly the same shape as the probability histogram?
Which scheme makes blind guessing worthless on average? Why does a penalty of exactly 1/3 do that on a four-choice test?
A student who knows some answers is not guessing at random. How would that change the model?
Modification for Distance Learning
Each student uses an online random number generator from 1 to 4 for the five answers and enters their number correct in a shared form, which builds the class dot plot automatically.
2
Distribution Stations
20 minGroups of 3-4
Groups rotate through four stations. At each one they state the model, develop the theoretical distribution of a random variable and find its expected value. Each station calls for a different counting method.
Station Cards
Station 1 (organized list): a spinner with three equal sections numbered 1, 2 and 3 is spun twice; X is the product of the two results
Station 2 (without replacement): a bag holds 3 red marbles and 1 green marble; two are drawn; X is the number of red marbles
Station 3 (two-stage process): flip a coin and roll a number cube; X is double the cube's number if the coin shows heads and the cube's number if it shows tails
Station 4 (tree): roll two number cubes; X is the number of cubes that show a prime number (2, 3 or 5)
At Station 3, some values of X can happen in two ways. Which ones, and how did you handle them?
At which station would a tree diagram be the least practical? Why?
3
Design a Fair Penalty
15-20 minPairs
Pairs design a grading rule for a four-question test with three choices per question so that a student who guesses on everything has an expected score of exactly 0. They develop the distribution first, then solve for the penalty.
Procedure
Develop the distribution of X, the number of correct guesses: 16, 32, 24, 8 and 1 out of 81 for 0 to 4 correct
Find E(X) = 108/81 = 4/3 and the expected number of wrong answers, 8/3
Let the rule be +1 per correct answer and -c per wrong answer. Solve 4/3 - (8/3)c = 0 to get c = 1/2
With c = 1/2, find the probability that a guesser still gets a positive score (2 or more correct: 33/81 = 11/27)
Discussion Questions
A fair penalty makes the expected score 0. Does it make every guesser score 0? What does your distribution say?
How would the fair penalty change for five choices per question? Test your idea.
Challenge Variation
Pairs write a rule that gives a guesser a negative expected score, and argue whether a teacher should use it.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Distribution of Correct Guesses on the Five-Question Test
The theoretical probability histogram of X, the number of correct answers when guessing on all five questions with four choices each, drawn to scale. Bar labels are the probabilities rounded to three decimals. The triangle marks the balance point, E(X) = 1.25.
Diagram 2: Tree Diagram for Two Guessed Questions
Each branch is labeled with its probability: C (correct) 1/4 and W (wrong) 3/4. Multiplying along a path gives the probability of that path, and paths with the same number of correct answers are added: P(X = 1) = 3/16 + 3/16 = 6/16.
04
Homework Assignment
~30 min
HSS.MD.A.3 Homework: Theoretical Distributions and Expected Value
Directions: For each problem, state the probability model, define the random variable, and show how you found each probability (a list, a tree diagram or a count of branches). Check that each distribution adds to 1 before finding the expected value.
Part 1: Developing Distributions (Problems 1-3)
A quiz has 3 multiple-choice questions with 5 choices each, and a student guesses on all of them. Let X be the number of correct answers. Draw a tree diagram, develop the probability distribution of X, and find E(X).
A bag holds 5 red tokens and 3 yellow tokens. Two tokens are drawn at random without replacement, and Y is the number of yellow tokens drawn. Develop the probability distribution of Y by counting pairs, and find E(Y).
Two fair number cubes are rolled. Let W be the smaller of the two numbers (the shared number if both are the same). Develop the probability distribution of W and find E(W) as a fraction and as a decimal to two places.
Part 2: Expected Scores and Grades (Problems 4-6)
A student guesses on every question of a 6-question true-false quiz. (a) Develop the distribution of X, the number of correct answers. (b) Find E(X). (c) The teacher gives 10 points per correct answer. What is the expected grade? (d) The student passes with 5 or more correct answers. What is the probability of passing?
A test has 4 questions with 5 choices each. It gives 1 point per correct answer and takes away 1/4 point per wrong answer. (a) Develop the distribution of the number of correct answers for a student who guesses on every question, and find the expected score. (b) Now the student can rule out one wrong choice on every question and guesses among the other 4. Find the new expected score.
Three students' graded papers are handed back at random, one to each student. Let M be the number of students who get their own paper. List the 6 equally likely ways to hand back the papers, develop the distribution of M, and find E(M). Explain why M = 2 is impossible.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Model and Variable
Model stated and variable defined
One of the two missing
Neither stated
Distribution
All branches or outcomes counted; adds to 1
Method correct, one value wrong
Branches missed or no distribution
Expected Value
Correct E(X) and expected score or grade
One arithmetic error
Missing or plain average of values
Interpretation
Explains results in context
Partly explained
No interpretation
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
A student guesses on 4 true-false questions. What is the probability that all 4 answers are correct?
Answer: C
Each guess is correct with probability 1/2, independently, so P(all 4) = (1/2)4 = 1/16. Choice A uses 1 out of 4 questions. Choice B multiplies only three of the four probabilities. Choice D is the chance for one question.
Question 2 of 20 · Multiple Choice
A student guesses on 3 multiple-choice questions with 4 choices each. What is the probability of exactly one correct answer?
Answer: A
One branch such as CWW has probability (1/4)(3/4)(3/4) = 9/64, and there are 3 such branches (CWW, WCW, WWC), so the probability is 27/64. Choice C counts only one branch. Choice B is the chance of a correct answer on a single question.
Question 3 of 20 · Multiple Choice
A student guesses on every question of a 20-question test with 4 choices per question. What is the expected number of correct answers?
Answer: D
Each question adds 1/4 of a correct answer on average, so E(X) = 20 × 1/4 = 5. Choice A divides by 5 instead of 4. Choice B treats each guess as a 50-50 chance. Choice C is 10 × 1/4, the value for a 10-question test.
Question 4 of 20 · Multiple Choice
Two fair number cubes are rolled, and X is the number of sixes. What is P(X = 1)?
Answer: B
Six on the first cube only: 1 × 5 = 5 outcomes; six on the second only: 5 more. So P(X = 1) = 10/36. Choice C counts the 11 outcomes with at least one six, which includes (6, 6), where X = 2. Choice D adds 1/6 + 1/6, which counts (6, 6) twice even though that outcome has X = 2.
Question 5 of 20 · Multiple Choice
A spinner lands on red with probability 0.3 and is spun 3 times. The number of reds X has P(X = 0) = 0.343, P(X = 1) = 0.441, P(X = 2) = 0.189 and P(X = 3) = 0.027. What is E(X)?
Answer: C
E(X) = 0(0.343) + 1(0.441) + 2(0.189) + 3(0.027) = 0.441 + 0.378 + 0.081 = 0.9, which also equals 3 × 0.3. Choice A is the plain mean of 0, 1, 2 and 3. Choice D is the most likely value, not the expected value.
Question 6 of 20 · Multiple Choice
A committee of 2 people is chosen at random from 3 teachers and 2 parents. What is the probability that both members are parents?
Answer: A
There are C(5, 2) = 10 equally likely pairs, and only one of them contains both parents, so the probability is 1/10. Choice C treats the three possible numbers of parents (0, 1 or 2) as equally likely. Choice D, (2/5)², would be correct only if the same person could be chosen twice.
Question 7 of 20 · Multiple Choice
On a 10-question test with 4 choices per question, a correct answer earns 10 points and a wrong answer loses 2 points. What is the expected score of a student who guesses on every question?
Answer: B
Expected correct: 10 × 1/4 = 2.5; expected wrong: 7.5. Expected score = 10(2.5) - 2(7.5) = 25 - 15 = 10 points. Choice A ignores the penalty. Choice D subtracts 2 points for only 2.5 wrong answers instead of 7.5.
Question 8 of 20 · Multiple Choice
Which of these is needed to develop a theoretical probability distribution?
Answer: D
Theoretical probabilities come from the structure of the process: fair cubes, random guesses, well-shuffled cards. Choices A, B and C all describe empirical probabilities estimated from data, which belong to HSS.MD.A.4.
Question 9 of 20 · Multiple Choice
A spinner lands on 1 with probability 1/2, on 2 with probability 1/4 and on 4 with probability 1/4. It is spun twice. What is the probability that the sum of the two spins is 5?
Answer: A
A sum of 5 needs a 1 and a 4 in either order: (1/2)(1/4) + (1/4)(1/2) = 1/8 + 1/8 = 1/4. Choice B counts only one order. Choice C treats the six possible sums (2, 3, 4, 5, 6, 8) as equally likely.
Question 10 of 20 · Multiple Choice
A fair number cube is rolled. X is the number rolled if it is even and 0 if it is odd. What is E(X)?
Answer: C
E(X) = (0 + 2 + 0 + 4 + 0 + 6)/6 = 12/6 = 2. Choice A is the expected value of the roll itself. Choice B averages only the even numbers 2, 4 and 6, which forgets the three outcomes where X = 0.
Question 11 of 20 · Multiple Choice
For the official example (guessing on all five questions of a test with four choices per question), what is the probability of 3 or more correct answers?
Answer: B
Add the probabilities of 3, 4 and 5 correct: (90 + 15 + 1)/1024 = 106/1024 ≈ 0.10. Choice A includes only exactly 3 correct. Choice D assumes that 3 or more is half of the six values, but the distribution is skewed right.
Question 12 of 20 · Multiple Choice
A fair coin is tossed 4 times. What is the probability of exactly 2 heads?
Answer: D
There are 24 = 16 outcomes and C(4, 2) = 6 of them have 2 heads, so the probability is 6/16 = 3/8. Choice A reasons that 2 is half of 4. Choice C treats the five possible values 0 to 4 as equally likely.
Question 13 of 20 · Multiple Choice
One card is drawn from a standard 52-card deck. An ace scores 10 points, a face card (jack, queen or king) scores 5 points, and any other card scores 0. What is the expected score?
Answer: A
E = 10(4/52) + 5(12/52) + 0(36/52) = (40 + 60)/52 = 100/52 = 25/13 ≈ 1.92 points. Choice D counts only the aces and choice C only the face cards. Choice B averages the three scores as if they were equally likely.
Question 14 of 20 · Multiple Choice
A student guesses on 4 questions with 4 choices each. What is the probability that none of the answers is correct?
Answer: C
Each answer is wrong with probability 3/4, independently, so P(none correct) = (3/4)4 = 81/256 ≈ 0.32. Choice B is the chance of missing one question. Choice D assumes that guessing on 4 questions guarantees at least one right.
Question 15 of 20 · Short Answer
A code is made of two digits, each chosen at random from 1, 2 and 3 (repeats allowed). Let X be the number of different digits in the code. Develop the probability distribution of X and find E(X).
There are 9 equally likely codes. Three of them (11, 22, 33) use one digit and the other six use two. P(X = 1) = 1/3, P(X = 2) = 2/3, and E(X) = 1(1/3) + 2(2/3) = 5/3 ≈ 1.67.
Question 16 of 20 · Short Answer
For the official example (five questions, four choices, guessing on all), a teacher uses a new scheme: 1 point per correct answer and minus 1/2 point per wrong answer. Find the expected score of a guesser.
The score is X - (5 - X)/2, so the expected score is E(X) - (5 - E(X))/2 = 1.25 - 3.75/2 = -0.625 points. Weighting each score by P(X = k) from the distribution gives the same result. The penalty is larger than 1/3, so guessing lowers the expected score.
Question 17 of 20 · Short Answer
A student says: "The expected number of correct answers when guessing on the five-question test is 1.25, so a guesser will get 1 or 2 answers right." Is the student right? Use the distribution to explain.
No. 1.25 is the mean over many guessers, not a promise for one guesser. From the distribution, P(X = 1) + P(X = 2) = (405 + 270)/1024 = 675/1024 ≈ 0.66, so about a third of guessers get 0 or 3 or more right. In particular P(X = 0) ≈ 0.24.
Question 18 of 20 · Short Answer
A box holds 7 light bulbs, 2 of which are defective. Three bulbs are chosen at random. Let D be the number of defective bulbs chosen. Develop the probability distribution of D and find E(D).
There are C(7, 3) = 35 equally likely groups. D = 0: C(5, 3) = 10 groups; D = 1: 2 × C(5, 2) = 20; D = 2: C(5, 1) = 5. P(D = 0) = 2/7, P(D = 1) = 4/7, P(D = 2) = 1/7, and E(D) = (20 + 10)/35 = 6/7 ≈ 0.86.
Question 19 of 20 · Short Answer
A student guesses on 2 multiple-choice questions with 5 choices each. Develop the probability distribution of X, the number of correct answers, and find E(X).
Each answer is right with probability 1/5. P(X = 0) = (4/5)² = 16/25, P(X = 1) = 2(1/5)(4/5) = 8/25, P(X = 2) = (1/5)² = 1/25. The sum is 25/25. E(X) = 8/25 + 2/25 = 10/25 = 0.4, which equals 2 × 1/5.
Question 20 of 20 · Short Answer
A fair four-sided die numbered 1 to 4 is rolled twice. Let X be the larger of the two results. Develop the probability distribution of X and find E(X).
Of the 16 outcomes, X = m for 2m - 1 of them: P(X = 1) = 1/16, P(X = 2) = 3/16, P(X = 3) = 5/16, P(X = 4) = 7/16. E(X) = (1 + 6 + 15 + 28)/16 = 50/16 = 3.125.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSS.MD.A.3 mean?
HSS.MD.A.3 means students can build a probability distribution for a random variable when the probabilities come from a model, not from data, and then find its expected value. Typical models are fair dice, shuffled cards, random draws and random guessing. The standard's own example is the number of correct answers when a student guesses on a five-question multiple-choice test.
What is a theoretical probability distribution?
It is a list of the values of a random variable with probabilities calculated from the structure of the chance process. For example, a fair number cube has six equally likely faces, so each face has probability 1/6 without anyone rolling it. An empirical distribution, the subject of HSS.MD.A.4, uses relative frequencies from observed data instead.
How do you find the probability distribution for guessing on a multiple-choice test?
Treat each question as an independent guess that is right with probability 1/c, where c is the number of choices. For n questions, a branch of the tree with k right answers has probability (1/c)k(1 - 1/c)n-k, and there are C(n, k) such branches. Multiply the two to get P(X = k), and repeat for every k from 0 to n.
What is the expected number of correct answers from guessing?
It is the number of questions divided by the number of choices. Each question contributes 1/c of a correct answer on average, so over n questions the expected number correct is n/c. Students can confirm this from the full distribution, as the lesson does for the five-question test.
Why do some tests take off points for wrong answers?
To make blind guessing worthless on average. On a four-choice test, a penalty of 1/3 point per wrong answer makes the expected score of a random guesser exactly 0, as the grading-scheme example shows. A student who can rule out some choices still gains on average by guessing, because their chance of a correct answer is higher than the model assumes.
Is HSS.MD.A.3 taught in Algebra 2 or Statistics?
It depends on the school. HSS.MD.A.3 is marked (+), the Common Core label for additional mathematics needed for advanced courses. It is usually taught in a statistics course or Precalculus, and some Algebra II courses include it alongside expected value.
Do students need combinations for HSS.MD.A.3?
Not always. Organized lists and tree diagrams are enough for small problems, such as two dice or three guessed questions. For larger ones, such as the five-question test, counting the branches with combinations, C(n, k), is much faster. Combinations are part of the (+) standard HSS.CP.B.9.
What is the difference between HSS.MD.A.3 and HSS.MD.A.4?
Both ask for a probability distribution and its expected value. In HSS.MD.A.3 the probabilities are theoretical: they follow from a model such as fair dice or random guessing. In HSS.MD.A.4 the probabilities are empirical: they come from data, such as the number of TV sets in a sample of households.
What mistakes do students make when developing a theoretical distribution?
A common one is counting only one branch for each value, for example treating "exactly one correct" as a single path when there are several. Others are adding probabilities along a branch instead of multiplying, treating the possible values as equally likely, and forgetting that drawing without replacement changes the probabilities at the second draw.
Is the guessing example a binomial distribution?
Yes. When there is a fixed number of independent trials, each with the same probability of success, the number of successes has a binomial distribution. The guessing test is one: 5 trials with success probability 1/4. HSS.MD.A.3 does not require the name, but students who continue in statistics will meet it.
07
Related Standards
6 standards
These standards connect to HSS.MD.A.3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSS.MD.A.2Prerequisite
Calculate the expected value of a random variable and interpret it as the mean
