Appendix A — Answer Key, Chapter 16: Probability: Theoretical and Experimental

SOL 7.PS.1 · Covers textbook Chapter 16 and the companion workbook. Item numbers match the textbook; workbook items are the same problems with the same numbers, so this key serves both. Reasoning answers show an acceptable response, not the only wording.

A note on the hands-on pages. Workbook pages 7, 8, and 13 ask students to roll a real die and flip a real coin. Those results will vary from student to student, and there is no single correct answer. For each of those activities this key gives the theoretical value to compare against, the expected frequency, and the range a reasonable result falls in. A student whose data land outside the usual range has not made an error — that is what variability looks like — but it is worth checking that the tally totals are right.

Lesson 16.1 — Theoretical Probability

Guided practice

1. $\{1, 2, 3, 4, 5, 6\}$; 6 outcomes.
2. $P(3) = \tfrac16$
3. $P(\text{odd}) = \tfrac36 = \tfrac12$
4. $P(B) = \tfrac14$, because the sector spans $90°$ of $360°$.
5. $P(\text{red}) = \tfrac{4}{12} = \tfrac13$
6. $P(\text{not red}) = 1 - \tfrac13 = \tfrac23$ (check by counting: $3 + 5 = 8$ marbles are not red, and $\tfrac{8}{12} = \tfrac23$).

Independent practice

7. a) $\tfrac16$ b) $\tfrac26 = \tfrac13$ (the 5 and the 6) c) $\tfrac06 = 0$ d) $\tfrac66 = 1$
8. Sample space: HH, HT, TH, TT. $P(\text{exactly one head}) = \tfrac24 = \tfrac12$; $P(\text{two tails}) = \tfrac14$.
9. $P(\text{red}) = \tfrac{180}{360} = \tfrac12$; $P(\text{blue}) = \tfrac{90}{360} = \tfrac14$; $P(\text{green or yellow}) = \tfrac{90}{360} = \tfrac14$. Sum: $\tfrac12 + \tfrac14 + \tfrac18 + \tfrac18 = \tfrac48 + \tfrac28 + \tfrac18 + \tfrac18 = \tfrac88 = 1$.
10. $P(\text{green}) = \tfrac{5}{12} \approx 0.417 \approx 41.7\%$; $P(\text{not green}) = \tfrac{7}{12} \approx 0.583 \approx 58.3\%$.
11. PROBABILITY has 11 letters. B appears twice, so $P(\text{B}) = \tfrac{2}{11}$. The vowels are O, A, I, I — four tiles — so $P(\text{vowel}) = \tfrac{4}{11}$.
12. Multiples of 3 among 1–8 are 3 and 6, so $P = \tfrac28 = \tfrac14$.
13. Probability $0$: any impossible event, such as rolling a $9$, rolling a $0$, or rolling a negative number. Probability $1$: any certain event, such as rolling a whole number from 1 to 6, or rolling a number less than 7.
14. Least to most likely: rolling a negative number ($0$), rolling a 2 ($\tfrac16$), rolling a number greater than 2 ($\tfrac46 = \tfrac23$), rolling a number greater than 0 ($1$).
15. $P(\text{Nadia wins}) = \tfrac{4}{25} = 0.16 = 16\%$
16. Every trial produces exactly one outcome from the sample space, so the favorable counts for all the separate outcomes add up to the total number of outcomes — and a number divided by itself is 1. On the unequal spinner the pointer must stop in some sector, so the four sector angles fill the whole circle: $180 + 90 + 45 + 45 = 360$, and $\tfrac{360}{360} = 1$.

Exit ticket 16.1

17. $\tfrac16$
18. $P(A \text{ or } C) = \tfrac24 = \tfrac12$
19. $P(\text{blue}) = \tfrac{3}{12} = \tfrac14$
20. A probability of $0$ means the event has no favorable outcomes and cannot happen; a probability of $1$ means every outcome is favorable, so the event must happen. For one die roll, $P(\text{rolling a }10) = 0$ and $P(\text{rolling a number less than }7) = 1$.

Lesson 16.2 — Experimental Probability

Guided practice

21. $9 + 12 + 8 + 11 + 10 + 10 = 60$. You check first because every experimental probability uses the total number of trials as its denominator. If the frequencies do not add to the number of trials, a count is wrong and every probability computed from the table would be wrong too.
22. $P_{\text{exp}}(1) = \tfrac{9}{60} = \tfrac{3}{20} = 0.15$
23. $P_{\text{exp}}(6) = \tfrac{10}{60} = \tfrac16 \approx 0.167$
24. $P_{\text{exp}}(\text{greater than }4) = \tfrac{10 + 10}{60} = \tfrac{20}{60} = \tfrac13 \approx 0.333$
25. $P_{\text{exp}}(\text{tails}) = \tfrac{17}{40} = 0.425$
26. $P_{\text{exp}}(C) = \tfrac{21}{80} = 0.2625$

Independent practice

27. $P_{\text{exp}}(2) = \tfrac{12}{60} = \tfrac15 = 0.20$; $P_{\text{exp}}(4) = \tfrac{11}{60} \approx 0.18$; $P_{\text{exp}}(5) = \tfrac{10}{60} = \tfrac16 \approx 0.17$.
28. Odd frequencies are $9 + 8 + 10 = 27$, so $P_{\text{exp}}(\text{odd}) = \tfrac{27}{60} = \tfrac{9}{20} = 0.45$. Since $P_{\text{exp}}(\text{even}) = \tfrac{33}{60} = \tfrac{11}{20} = 0.55$, the sum is $\tfrac{9}{20} + \tfrac{11}{20} = \tfrac{20}{20} = 1$. It had to happen because every one of the 60 rolls was either odd or even, so the two frequencies add to all 60 trials.
29. $P_{\text{exp}}(B) = \tfrac{17}{80} = 0.2125$; $P_{\text{exp}}(D) = \tfrac{18}{80} = \tfrac{9}{40} = 0.225$; $P_{\text{exp}}(B \text{ or } D) = \tfrac{17 + 18}{80} = \tfrac{35}{80} = \tfrac{7}{16} = 0.4375$.
30. $14 + 11 + 13 = 38$, so D occurred $50 - 38 = 12$ times. $P_{\text{exp}}(D) = \tfrac{12}{50} = \tfrac{6}{25} = 0.24$.
31. $P_{\text{exp}}(\text{point up}) = \tfrac{33}{60} = \tfrac{11}{20} = 0.55$; $P_{\text{exp}}(\text{point down}) = \tfrac{27}{60} = \tfrac{9}{20} = 0.45$. Theoretical probability cannot be used because the two outcomes are not equally likely and there is no way to reason out how likely each one is — the answer depends on the shape, weight, and balance of that particular tack, so it can only be measured.
32. $P_{\text{exp}}(\text{make}) = \tfrac{12}{20} = \tfrac35 = 0.6$; expect about $0.6 \times 100 = 60$ makes in 100 attempts.
33. $P_{\text{exp}}(\text{prize}) = \tfrac{30}{250} = \tfrac{3}{25} = 0.12 = 12\%$
34. $P_{\text{exp}}(\text{scratched}) = \tfrac{16}{400} = \tfrac{1}{25} = 0.04$; expect about $0.04 \times 2500 = 100$ scratched cases.
35. Experimental probability is a summary of things that actually happened, so there is nothing to summarize until the trials are run. Theoretical probability comes from counting possibilities, which you can do in your head; experimental probability comes from counting results, which only exist after you collect them.
36. $\tfrac{9}{20} = 0.45$ and $\tfrac{13}{20} = 0.65$. Both are correct, because each student is reporting what actually happened in their own 20 trials. Twenty flips is a small number of trials, so ordinary variability easily produces results this far apart, and neither student made a mistake.

Exit ticket 16.2

37. $P_{\text{exp}}(\text{red}) = \tfrac{12}{30} = \tfrac25 = 0.4$
38. $P_{\text{exp}}(5) = \tfrac{10}{60} = \tfrac16 \approx 0.167$
39. $P_{\text{exp}}(\text{heads}) = \tfrac{23}{40} = 0.575$
40. No. The frequencies would almost certainly come out different, because each roll is random and 60 rolls is a small number of trials. What stays the same is the die itself and therefore the theoretical probability of each face, $\tfrac16$. The new experimental probabilities would again land somewhere near $\tfrac16$ without matching it.

Lesson 16.3 — What Happens as the Number of Trials Grows

Guided practice

41. $\tfrac{54}{100} = \tfrac{27}{50} = 0.54$
42. At 20 flips: $|0.55 - 0.5| = 0.05$. At 500 flips: $|0.51 - 0.5| = 0.01$. The 500-flip result is closer.
43. No. The distance grew twice: from $0.000$ at 10 flips to $0.050$ at 20 flips, and again from $0.050$ at 20 flips to $0.060$ at 50 flips.
44. $P(\text{heads}) = \tfrac12$. The flips are independent, so the six earlier heads change nothing.
45. At 100 flips, half of 100 is 50 and there were 54 heads, so the gap is $|54 - 50| = 4$. At 1{,}000 flips, half is 500 and there were 496 heads, so the gap is $|496 - 500| = 4$.
46. "…the experimental probability tends to get closer to the theoretical probability." Wordings such as "usually gets closer to" or "is likely to get closer to" are also correct. Answers of "equals" or "becomes the same as" are not correct.

Independent practice

47. $\tfrac{255}{500} = \tfrac{51}{100} = 0.51$
48. Distances: 20 flips $\to 0.05$; 100 flips $\to 0.04$; 1{,}000 flips $\to 0.004$. Farthest to closest: 20 flips, 100 flips, 1{,}000 flips.
49. The table's 1{,}000-flip row gives $\tfrac{496}{1000} = 0.496$, not $0.5$, so his claim is false for this simulation. A correct version: "If you flip a fair coin 1{,}000 times, the experimental probability of heads will probably be close to $0.5$, but it will almost certainly not equal $0.5$ exactly."
50. Spins are independent: the spinner has no memory of the five blue results, and nothing about it changes between spins. The sector sizes are the only thing that determines the probability, and they did not change, so the probability of blue on the next spin is exactly what it was on the first spin. Priya is committing the gambler's fallacy.
51. $P_{\text{exp}}(6) = \tfrac{15}{60} = \tfrac14 = 0.25$. After 600 rolls you would expect the experimental probability to be closer to the theoretical $\tfrac16 \approx 0.167$ than $0.25$ is. That is a reasonable expectation, not a guarantee — a larger sample makes a close result much more likely without making it certain.
52. The estimate from 3{,}000 trials, because experimental probabilities from large samples vary much less and tend to sit closer to the true value. It is not guaranteed to be closer; it is only much more likely to be.
53. Short test: $\tfrac{8}{20} = \tfrac25 = 0.4$. Long test: $\tfrac{46}{200} = \tfrac{23}{100} = 0.23$. The long test is the better estimate, because 200 trials is far less affected by ordinary variability than 20 trials.
54. $P_{\text{exp}}(6) = \tfrac{97}{600} \approx 0.162$, compared with $\tfrac16 \approx 0.167$. The difference is only about $0.005$, and the expected frequency was $\tfrac16 \times 600 = 100$ against an actual 97. Nothing here suggests the die is unfair; being off by 3 rolls in 600 is exactly the kind of variability a fair die produces.
55. It means the outcome of a flip does not depend in any way on earlier flips — the coin carries no record of what it did, so nothing is ever "due." This does not contradict the drift toward $0.5$, because that drift is not caused by later flips correcting earlier ones. Early lopsided results are never cancelled; they are simply outweighed as more and more flips are added to the fraction.
56. At 100 flips the head count was 4 away from half, and at 1{,}000 flips it was still 4 away — yet the experimental probability moved from $0.54$ to $0.496$. A gap of 4 is a large share of 100 trials ($0.04$) and a tiny share of 1{,}000 trials ($0.004$). The ratio shrinks because the denominator grows much faster than the gap does, even in runs where the gap itself grows.

Exit ticket 16.3

57. The 1{,}000-flip row, with $P_{\text{exp}} = 0.496$ and a distance of $0.004$ from $0.5$.
58. $\tfrac12$. Each flip is independent, so four tails in a row change nothing.
59. False. More trials make the experimental probability tend toward $\tfrac16$, but no number of rolls makes it equal $\tfrac16$ exactly and stay there. In fact, for most numbers of trials it is impossible: 100 rolls cannot produce an experimental probability of exactly $\tfrac16$, since $\tfrac{100}{6}$ is not a whole number of rolls.
60. Acceptable: "As the number of trials increases, the experimental probability tends to get closer to the theoretical probability, though it can move farther away at any particular step and never lands on it exactly." Answers using "always equals" or "eventually becomes" are not acceptable.

Lesson 16.4 — Comparing Theoretical and Experimental Probability

Guided practice

61. $P(3) = \tfrac16 \approx 0.167$; $P_{\text{exp}}(3) = \tfrac{8}{60} = \tfrac{2}{15} \approx 0.133$. Difference: $\tfrac16 - \tfrac{2}{15} = \tfrac{10}{60} - \tfrac{8}{60} = \tfrac{2}{60} = \tfrac{1}{30} \approx 0.033$.
62. $\tfrac16 \times 60 = 10$
63. $12 - 10 = 2$ more than expected.
64. $P(\text{even}) = \tfrac12 = 0.5$; $P_{\text{exp}}(\text{even}) = \tfrac{33}{60} = \tfrac{11}{20} = 0.55$. The experimental value is $0.05$ higher.
65. Expected frequency $= \tfrac12 \times 40 = 20$; the actual 23 is 3 more than expected.
66. Expected frequency $= \tfrac14 \times 80 = 20$; Ana's 24 is 4 more than expected.

Independent practice

| Outcome | 1 | 3 | 4 |
|---|---|---|---|
| Frequency out of 60 | 9 | 8 | 11 |
| $P_{\text{exp}}$ as a fraction | $\tfrac{9}{60} = \tfrac{3}{20}$ | $\tfrac{8}{60} = \tfrac{2}{15}$ | $\tfrac{11}{60}$ |
| $P_{\text{exp}}$ as a decimal | $0.150$ | $0.133$ | $0.183$ |
| Difference from $\tfrac16$ | $-\tfrac{1}{60} \approx -0.017$ | $-\tfrac{1}{30} \approx -0.033$ | $+\tfrac{1}{60} \approx +0.017$ |

Accept differences computed against the rounded $0.167$, which give $-0.017$, $-0.034$, and $+0.016$.

68. $P(\text{greater than }4) = \tfrac26 = \tfrac13$ and $P_{\text{exp}}(\text{greater than }4) = \tfrac{20}{60} = \tfrac13$. They are exactly equal. That tells you this particular set of 60 rolls happened to split evenly for this event — a pleasant coincidence. It does not tell you the experiment was run especially well, that the die is definitely fair, or that repeating the 60 rolls would match again.
69. Expected frequency $= \tfrac12 \times 500 = 250$; the actual 255 is 5 more than expected.
70. $P(\text{green}) = \tfrac{5}{12}$, so the expected frequency is $\tfrac{5}{12} \times 120 = 50$.
71. Set the generator to produce whole numbers from 1 to 4, each equally likely, and run it 100 times. Let 1 stand for sector A, 2 for B, 3 for C, and 4 for D. Record the resulting counts as if they were spins. This works because the four sectors are equal, so the four spinner outcomes are equally likely, just like the four numbers.
72. $P_{\text{exp}}(6) = \tfrac{34}{60} = \tfrac{17}{30} \approx 0.567$, against a theoretical $\tfrac16 \approx 0.167$ and an expected frequency of 10. Getting 34 sixes where 10 were expected is far outside ordinary variability, so the die is very likely not fair — most likely weighted toward the 6.
73. Devon is missing variability. Random results are naturally lumpy, so even a perfectly run experiment with a perfectly fair die will rarely produce frequencies that match the expected ones exactly. Small differences are the normal state of affairs, not evidence of a counting error. Only a very large difference is worth investigating.
74. Expected wins $= \tfrac15 \times 300 = 60$; $P_{\text{exp}}(\text{win}) = \tfrac{41}{300} \approx 0.137$. The club won 19 fewer times than advertised out of 300 plays, and $0.137$ is noticeably below $0.2$. That gap is large enough to be worth questioning, so the advertised rate looks too high — though a single investigation is a reason to test further, not proof on its own.
75. Expected prizes $= \tfrac16 \times 90 = 15$; $P_{\text{exp}}(\text{prize}) = \tfrac{14}{90} = \tfrac{7}{45} \approx 0.156$. Finding 14 prizes where 15 were expected is a difference of 1 box, well within ordinary variability, so this result is consistent with the company's claim.
76. Two reasons: (1) Variability — random outcomes are lumpy, so any finite set of trials will land near, but not on, the expected split. (2) Too few trials — small samples swing widely, so a short investigation can produce an experimental probability far from the theoretical one purely by chance. A third acceptable reason: the theoretical model may not describe the real object, as with a weighted die or unequal spinner sectors.
77. Compare the actual frequency with the expected frequency and ask how big the gap is relative to the number of trials. A gap of a few rolls out of 60 is ordinary. A gap like 34 sixes where 10 were expected is not something chance produces, so it points to the model being wrong. It also helps to run more trials: ordinary variability shrinks as a share of the total, while a genuinely wrong model keeps producing the same lopsided results no matter how many trials you add.
78. Yes. For a fair die $P(6) = \tfrac16 > 0$, but if you roll it 10 times and no 6 appears, then $P_{\text{exp}}(6) = \tfrac{0}{10} = 0$. An experimental probability of 0 only means the event did not occur in those trials; it does not mean the event is impossible.

Exit ticket 16.4

79. $P(1) = \tfrac16 \approx 0.167$; $P_{\text{exp}}(1) = \tfrac{9}{60} = \tfrac{3}{20} = 0.15$.
80. Expected frequency $= \tfrac12 \times 200 = 100$; the record shows 96 heads, which is 4 fewer than expected — a small difference for 200 flips.
81. $P(A) = \tfrac14 = 0.25$ and $P_{\text{exp}}(A) = \tfrac{10}{40} = \tfrac14 = 0.25$. They happen to be exactly equal in this set of 40 spins, which is a coincidence rather than a guarantee about future spins.
82. Theoretical probability is found by reasoning about the sample space — counting favorable outcomes out of equally likely total outcomes — and it does not change no matter how many trials you run. Experimental probability is found by running trials and dividing the number of times the event happened by the number of trials, so it changes every time you repeat the investigation. Only experimental probability can be found when the outcomes are not equally likely or the sample space is unknown, such as the probability that a thumbtack lands point up.

Chapter 16 Review

Part A — Theoretical probability (7.PS.1a)

83. $P(3) = \tfrac16$; $P(\text{odd}) = \tfrac36 = \tfrac12$; $P(\text{less than }3) = \tfrac26 = \tfrac13$.
84. $P(\text{red}) = \tfrac{4}{12} = \tfrac13$; $P(\text{blue}) = \tfrac{3}{12} = \tfrac14$; $P(\text{not green}) = \tfrac{7}{12}$.
85. $P(\text{red}) = \tfrac{180}{360} = \tfrac12$; $P(\text{green or yellow}) = \tfrac{90}{360} = \tfrac14$.
86. Sample space: HH, HT, TH, TT. $P(\text{two heads}) = \tfrac14$; $P(\text{at least one head}) = \tfrac34$.
87. Probability $0$: rolling a 9 (or any number not on the die). Probability $1$: rolling a number less than 7 (or any whole number from 1 to 6).
88. Greater than 5 means 6, 7, or 8, so $P = \tfrac38$.
89. $P(\text{not } E) = 1 - \tfrac27 = \tfrac57$.

Part B — Experimental probability (7.PS.1b)

Check first: $14 + 19 + 16 + 17 + 18 + 16 = 100$.

90. $P_{\text{exp}}(2) = \tfrac{19}{100} = 0.19$
91. $P_{\text{exp}}(6) = \tfrac{16}{100} = \tfrac{4}{25} = 0.16$
92. Even frequencies: $19 + 17 + 16 = 52$, so $P_{\text{exp}}(\text{even}) = \tfrac{52}{100} = \tfrac{13}{25} = 0.52$.
93. Odd frequencies: $14 + 16 + 18 = 48$, so $P_{\text{exp}}(\text{odd}) = \tfrac{48}{100} = \tfrac{12}{25} = 0.48$. Sum: $\tfrac{13}{25} + \tfrac{12}{25} = \tfrac{25}{25} = 1$.
94. Less than 3 means 1 or 2: $14 + 19 = 33$, so $P_{\text{exp}}(\text{less than }3) = \tfrac{33}{100} = 0.33$.
95. $P_{\text{exp}}(\text{heads}) = \tfrac{37}{80} = 0.4625$; $P_{\text{exp}}(\text{tails}) = \tfrac{43}{80} = 0.5375$.

Part C — Changes as the number of trials increases (7.PS.1c)

96. At 20 flips: $|0.55 - 0.5| = 0.05$. At 1{,}000 flips: $|0.496 - 0.5| = 0.004$.
97. The 1{,}000-flip result. The 10-flip result was exactly $0.5$, but that was luck, not accuracy — with only 10 flips the experimental probability can only take the values $0, 0.1, 0.2, \ldots, 1$, and it swings enormously from one set of 10 flips to the next. The 1{,}000-flip result of $0.496$ is farther from $0.5$ than the 10-flip result was, and it is still the far more reliable estimate, because estimates from large samples vary much less.
98. $\tfrac12$. The flips are independent, so the run of seven heads has no effect.
99. False. The experimental probability tends toward $0.5$ but never settles on it exactly; the 1{,}000-flip row in the table gives $0.496$. Adding more flips keeps moving it.
100. Acceptable: "As the number of trials increases, the experimental probability tends to get closer to the theoretical probability, but it can move farther away at individual steps and never becomes exactly equal to it."

Part D — Comparing experimental and theoretical probability (7.PS.1d)

101. $P(3) = \tfrac16 \approx 0.167$; $P_{\text{exp}}(3) = \tfrac{16}{100} = \tfrac{4}{25} = 0.16$. Difference: $\tfrac16 - \tfrac{4}{25} = \tfrac{25}{150} - \tfrac{24}{150} = \tfrac{1}{150} \approx 0.007$.
102. Expected frequency $= \tfrac16 \times 100 = \tfrac{100}{6} \approx 16.7$. Distances from $16.7$: outcome 1 is $2.7$; outcome 2 is $2.3$; outcome 3 is $0.7$; outcome 4 is $0.3$; outcome 5 is $1.3$; outcome 6 is $0.7$. The farthest is outcome 1, with 14 rolls.
103. Expected frequency $= \tfrac12 \times 80 = 40$; the actual 37 is 3 fewer than expected. That is a small difference for 80 flips and is consistent with a fair coin.
104. Expected frequency $= \tfrac16 \times 120 = 20$; $P_{\text{exp}}(6) = \tfrac{42}{120} = \tfrac{7}{20} = 0.35$. The die is likely not fair: 42 sixes where 20 were expected is more than double the expectation, a gap far too large for ordinary variability over 120 rolls.
105. Use a source of random digits. Read one digit at a time. Treat the digits 1, 2, 3, 4, 5, and 6 as the corresponding die rolls, and skip the digits 7, 8, 9, and 0 entirely, moving to the next digit without recording anything. Continue until you have recorded 100 rolls. Skipping is essential: if you assigned 7, 8, 9, and 0 to faces, those faces would come up more often than the others and the six simulated outcomes would no longer be equally likely.

Part E — Mixed application and reasoning

106. $P(\text{Trey wins}) = \tfrac{10}{250} = \tfrac{1}{25} = 0.04 = 4\%$; $P(\text{Trey does not win}) = 1 - \tfrac{1}{25} = \tfrac{24}{25} = 0.96 = 96\%$.
107. Check: $191 + 108 + 45 + 56 = 400$. Expected frequencies: red $\tfrac12 \times 400 = 200$; blue $\tfrac14 \times 400 = 100$; green $\tfrac18 \times 400 = 50$; yellow $\tfrac18 \times 400 = 50$. Differences: $-9$, $+8$, $-5$, $+6$, which add to $0$ as they must. $P_{\text{exp}}(\text{red}) = \tfrac{191}{400} = 0.4775$, close to the theoretical $0.5$. Yes, the results are consistent with the stated sector sizes: every color is within 9 spins of its expected frequency out of 400 trials, which is ordinary variability.
108. A probability is the number of favorable outcomes divided by the total number of outcomes. The count of favorable outcomes can never be less than 0, so the fraction can never be negative — ruling out $-0.2$. And the favorable outcomes are part of the total, so the numerator can never exceed the denominator, which means the fraction can never exceed 1 — ruling out $1.4$. Every probability satisfies $0 \le P \le 1$.
109. Spins are independent: the spinner's sectors do not change between spins, and the spinner has no memory of the four reds. Nothing about the past four spins makes red any less likely now. This is the gambler's fallacy. The correct probability of red on the next spin is $\tfrac12$, exactly as it was on the first spin.
110. Theoretical probability is computed by reasoning about the sample space — favorable outcomes divided by total equally likely outcomes — and it stays fixed. Experimental probability is computed from the record of an actual investigation — times the event occurred divided by number of trials — and it changes each time the investigation is repeated. Only experimental probability can be determined when the outcomes are not equally likely or the sample space is unknown: for example, the probability that a particular thumbtack lands point up, that a battery from a factory line is defective, or that a specific basketball player makes a free throw.

Workbook-only items

Page 2, fill in the blanks. The complete list of every possible result is the sample space. One single possible result is an outcome. An outcome you are counting for your event is called a favorable outcome.

Page 2, event table and scale.

Page 3, equal spinner. $P(A) = P(B) = P(C) = P(D) = \tfrac14$.

Page 3, unequal spinner table.

Page 3, watch out. Because the four sectors are not the same size, the four colors are not equally likely, so you cannot divide 1 by the number of colors. Blue covers $90°$ of the $360°$ circle, which is $\tfrac14$ here by coincidence of its angle — but red covers $180°$, so $P(\text{red}) = \tfrac12$, not $\tfrac14$. The probability comes from the angle, never from the number of labels.

Page 4, two coins. $P(\text{two heads}) = \tfrac14$; $P(\text{exactly one head}) = \tfrac12$; $P(\text{at least one head}) = \tfrac34$.

Page 7, Investigation 1 (roll a die 60 times). Results vary. Every student's table will be different, and that is the point of the activity. What must be true of any correct page: the six frequencies add to exactly 60, each experimental probability is that frequency over 60, and the six experimental probabilities add to 1. For comparison, the theoretical probability of each face is $\tfrac16 \approx 0.167$ and the expected frequency of each face is 10. In 60 rolls of a fair die, individual face counts between about 5 and 15 are entirely ordinary; $P_{\text{exp}}(\text{even})$ will usually land between about $0.35$ and $0.65$, against a theoretical $\tfrac12$. A count of, say, 25 for one face would be unusual enough to be worth re-rolling and re-counting.

Page 8, Investigation 2 (flip a coin 40 times). Results vary. Requirements for a correct page: heads plus tails equals exactly 40, and the two experimental probabilities add to 1 because every flip was either heads or tails. The theoretical probability of heads is $\tfrac12$ and the expected frequency is 20. In 40 flips, head counts from about 14 to 26 are entirely ordinary. Comparing with a classmate: the two experimental probabilities will usually differ, and both records are correct — 40 trials is a small sample, and different samples give different results without anyone making a mistake.

Page 9, class data. These are the same figures used in the textbook and are already filled in: 60 rolls with frequencies 9, 12, 8, 11, 10, 10; 40 flips with 23 heads and 17 tails; 80 spins with A 24, B 17, C 21, D 18.

Page 12, table completion.

Page 12, circle the correct wording. "tends to get closer to"; "still $\tfrac12$ heads".

Page 12, key sentence. Each trial is independent. The coin has no memory, so no outcome is ever "due."

Page 13, Investigation 3 (running record of 50 flips). Results vary. Requirements for a correct page: the five block counts add to the 50-flip total, each running total is the sum of all blocks so far (so the running totals never decrease), and each $P_{\text{exp}}$ is the running head count over the running flip count.

• Did the distance from $0.5$ shrink at every step? Usually no, and for many students it will grow at least once. That is the expected answer and nothing went wrong. The tendency toward $0.5$ shows up over many trials, not at every checkpoint, and 50 flips is still a small investigation.
• Class combination. Pooling the class gives a much larger number of trials, so the class $P_{\text{exp}}$ is usually closer to $0.5$ than an individual student's 50-flip result. It is not guaranteed to be closer — a larger sample makes a close result much more likely, not certain. For reference, with 25 students the class total is 1{,}250 flips, and a class result between about $0.47$ and $0.53$ would be entirely ordinary.

Page 16, comparison table.

Page 16, expected-frequency table. Expected frequency is 10 for every outcome, totaling 60. Differences (actual minus expected): $-1$, $+2$, $-2$, $+1$, $0$, $0$, which add to $0$.

Page 21, item 107 table. Expected frequencies: red 200, blue 100, green 50, yellow 50. Differences (actual minus expected): red $-9$, blue $+8$, green $-5$, yellow $+6$; these add to $0$.