Appendix A — Answer Key, Chapter 15: Probability of Independent and Dependent Events

SOL 8.PS.1 · Covers textbook Chapter 15 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 130 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Every probability in this key has been recomputed from its counts, and every fraction is given in lowest terms. For each complete set of four outcomes, the four probabilities have been checked to sum to exactly $1$. In every dependent-event answer, both the numerator and the denominator of the second factor were re-derived from what is left in the collection, because reducing only the denominator is the defining error of this topic.

Notation note: branch fractions inside a tree are shown unreduced, since $\frac{3}{9}$ records "three green left out of nine counters left." Final answers are reduced.

Lesson 15.1 — Independent or Dependent: What Replacement Changes

Guided practice

1. $5$ marbles.
2. $5$ marbles. The marble was put back, so the bag is restored.
3. $4$ marbles. The marble was kept out.
4. $2$ red and $2$ blue, total $4$. Both the red count and the total dropped by one.
5. $3$ red and $1$ blue, total $4$. Only the blue count dropped; red is untouched.
6. Independent. A coin cannot change the six faces of a die.
7. Dependent. The second draw comes from a bag with one fewer marble in it.
8. Replacement restores the collection, so the second event draws from exactly what the first event drew from.

Independent practice

9. a) independent b) dependent c) independent d) independent e) dependent
10. $4$ white and $6$ black, total $10$ — unchanged, because the tile was replaced.
11. $3$ white and $6$ black, total $9$.
12. $4$ white and $5$ black, total $9$.
13. First denominator $7$; second denominator $6$.
14. With replacement: $52$ and $52$. Without replacement: $52$ and $51$.
15. Putting the item back restores every count, so the second draw faces the same favorable count over the same total. Nothing about the first result changes the second probability, which is the definition of independence.
16. If the first was red, $4$ of the $9$ remaining are red: $P = \frac{4}{9}$. If the first was green, $5$ of the $9$ remaining are red: $P = \frac{5}{9}$. The second is larger. The probability of "second is red" has two different values depending on the first result, and that is exactly what dependent means.
17. Dependent. The button is not returned, so the second draw comes from $11$ buttons, and the count of the dropped button's type is one lower.
18. Independent. Two separate spinners; spinning one does not remove or add a sector on the other, so nothing about the first spin changes the second spin's probabilities.
19. Yes. Choosing two different students from a class for two jobs is dependent even though nothing is in a bag: once a student is chosen for the first job, that student is unavailable, so the second probability has a denominator one smaller. Any situation in which the first result removes a possibility is dependent.
20. The student is testing the results rather than the setup. Color has nothing to do with it. The test for independence is whether the first event changes the probability of the second — here, whether the marble was replaced.

Exit ticket 15.1

21. Two events are independent when knowing how the first turned out would not change your answer for the second.
22. With replacement, $10$. Without replacement, $9$.
23. $5$ yellow and $4$ purple, total $9$.
24. Independent. The chip is put back, so the second draw comes from the same full collection.

Lesson 15.2 — The Probability of Two Independent Events

Guided practice

25. $\frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12}$
26. $\frac{1}{2} \cdot \frac{3}{6} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$
27. The shaded outcomes are $H2$, $H4$, $H6$, so $\frac{3}{12} = \frac{1}{4}$. Yes, it matches $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.
28. $\frac{3}{5} \cdot \frac{3}{5} = \frac{9}{25}$
29. $\frac{2}{5} \cdot \frac{2}{5} = \frac{4}{25}$
30. $\frac{3}{5} \cdot \frac{2}{5} = \frac{6}{25}$
31. $\frac{9}{25} + \frac{6}{25} + \frac{6}{25} + \frac{4}{25} = \frac{25}{25} = 1$. The four outcomes are the only possibilities, and something must happen, so their probabilities must total $1$.
32. $\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$

Independent practice

33. $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$
34. $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$
35. $P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$ on each die, so $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.
36. $\frac{1}{5} \cdot \frac{1}{5} = \frac{1}{25}$
37. The even sectors are $2$ and $4$, so $\frac{2}{5} \cdot \frac{1}{2} = \frac{2}{10} = \frac{1}{5}$.
38. $\frac{2}{5} \cdot \frac{2}{5} = \frac{4}{25}$
39. $\frac{4}{10} \cdot \frac{6}{10} = \frac{24}{100} = \frac{6}{25}$
40. $\frac{6}{10} \cdot \frac{6}{10} = \frac{36}{100} = \frac{9}{25} = 0.36 = 36\%$
41. The primes on a die are $2$, $3$, and $5$, so $P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$ and $\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$.
42. $P(\text{no heads}) = P(\text{tails and tails}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$, so $P(\text{at least one head}) = 1 - \frac{1}{4} = \frac{3}{4}$.
43. Two boxes from a huge randomly mixed shipment are independent: $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$.
44. $\frac{2}{5} \cdot \frac{2}{5} = \frac{4}{25} = 0.16 = 16\%$
45. $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$. Multiplying by a fraction less than $1$ makes a quantity smaller, so requiring both events cannot be more likely than requiring either one alone. Demanding more is always harder.
46. The student added instead of multiplying. Adding is for a single event with two acceptable outcomes ("heads or tails"), not for two events that must both happen. The correct answer is $\frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12}$. Notice also that $\frac{2}{3}$ is larger than $\frac{1}{2}$, which is impossible for "both."

Exit ticket 15.2

47. $\frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12}$
48. $\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$
49. $\frac{3}{5} \cdot \frac{3}{5} = \frac{9}{25} = 0.36 = 36\%$
50. $P(A \text{ and } B) = P(A) \cdot P(B)$

Lesson 15.3 — The Probability of Two Dependent Events

Guided practice

51. $\frac{3}{5} \cdot \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$. The second factor is $\frac{2}{4}$, not $\frac{3}{4}$: a red marble is gone, so only $2$ red remain of $4$ marbles.
52. $\frac{3}{5} \cdot \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$. Here blue is untouched — still $2$ blue, but of $4$ marbles.
53. $\frac{2}{5} \cdot \frac{1}{4} = \frac{2}{20} = \frac{1}{10}$
54. $\frac{3}{10} + \frac{3}{10} + \frac{3}{10} + \frac{1}{10} = \frac{10}{10} = 1$. The four outcomes cover every possibility, so the tree is complete and consistent.
55. $\frac{4}{10} \cdot \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$
56. $\frac{6}{10} \cdot \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$
57. $\frac{4}{10} \cdot \frac{6}{9} = \frac{24}{90} = \frac{4}{15}$
58. $P(YG) = \frac{6}{10} \cdot \frac{4}{9} = \frac{24}{90} = \frac{4}{15}$, so with a common denominator of $15$: $\frac{2}{15} + \frac{4}{15} + \frac{4}{15} + \frac{5}{15} = \frac{15}{15} = 1$.

Independent practice

59. $\frac{5}{8} \cdot \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$
60. $\frac{3}{8} \cdot \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$
61. $\frac{5}{8} \cdot \frac{3}{7} = \frac{15}{56}$
62. $\frac{3}{8} \cdot \frac{5}{7} = \frac{15}{56}$. The same as item 61, which is worth noticing: order does not change the product.
63. Using $56$ as the common denominator: $\frac{20}{56} + \frac{6}{56} + \frac{15}{56} + \frac{15}{56} = \frac{56}{56} = 1$.
64. $\frac{4}{52} \cdot \frac{3}{51} = \frac{1}{13} \cdot \frac{1}{17} = \frac{1}{221}$
65. $\frac{13}{52} \cdot \frac{12}{51} = \frac{1}{4} \cdot \frac{4}{17} = \frac{1}{17}$
66. $\frac{6}{10} \cdot \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$
67. $\frac{4}{10} \cdot \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$
68. $\frac{1}{3} + \frac{2}{15} = \frac{5}{15} + \frac{2}{15} = \frac{7}{15}$. Two separate ways to match, so the two probabilities are added.
69. Two different students, so this is without replacement: $\frac{8}{20} \cdot \frac{7}{19} = \frac{56}{380} = \frac{14}{95}$.
70. $\frac{3}{15} \cdot \frac{2}{14} = \frac{1}{5} \cdot \frac{1}{7} = \frac{1}{35}$
71. $\frac{4}{9} \cdot \frac{5}{8} = \frac{20}{72} = \frac{5}{18}$. The oranges are untouched by eating an apple — still $5$ — but the total is now $8$.
72. The student reduced the total from $8$ to $7$ correctly in spirit but wrote $8$ again in the denominator, and used $4$ red over $8$ instead of $4$ red over $7$. Without replacement the second denominator must be $7$: $\frac{5}{8} \cdot \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$.

Exit ticket 15.3

73. $\frac{7}{10} \cdot \frac{6}{9} = \frac{42}{90} = \frac{7}{15}$
74. $\frac{3}{10} \cdot \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$
75. $\frac{7}{10} \cdot \frac{3}{9} = \frac{21}{90} = \frac{7}{30}$
76. The second factor must be recounted from what is left: the total drops by one, and the count of the kind already taken drops by one as well.

Lesson 15.4 — Comparing and Contrasting the Two Cases

Guided practice

77. With replacement $\frac{9}{25} = 0.36$; without replacement $\frac{3}{10} = 0.30$. With replacement is larger.
78. With replacement $\frac{4}{25} = 0.16$; without replacement $\frac{1}{10} = 0.10$. With replacement is larger.
79. With replacement $\frac{6}{25} = 0.24$; without replacement $\frac{3}{10} = 0.30$. Without replacement is larger.
80. Getting the same color twice needs that color to still be well represented on the second draw. Removing one of them lowers that color's count and the total, and the count falls by a larger share, so the second factor shrinks.
81. Removing one marble of the first color leaves the other color's count unchanged while the total drops, so the other color is a larger share of what remains and the second factor grows.
82. The first factor. Nothing has been drawn yet when the first event happens, so the collection is the original one in both cases.
83. With replacement the second denominator is the same as the first; without replacement it is one less than the first.

Independent practice

84. With: $\frac{4}{10} \cdot \frac{4}{10} = \frac{16}{100} = \frac{4}{25} = 0.16$. Without: $\frac{4}{10} \cdot \frac{3}{9} = \frac{12}{90} = \frac{2}{15} \approx 0.133$. With replacement is larger.
85. With: $\frac{6}{10} \cdot \frac{6}{10} = \frac{36}{100} = \frac{9}{25} = 0.36$. Without: $\frac{6}{10} \cdot \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \approx 0.333$. With replacement is larger.
86. With: $\frac{4}{10} \cdot \frac{6}{10} = \frac{6}{25} = 0.24$. Without: $\frac{4}{10} \cdot \frac{6}{9} = \frac{24}{90} = \frac{4}{15} \approx 0.267$. Without replacement is larger, because the colors differ.
87. With: $\frac{5}{8} \cdot \frac{5}{8} = \frac{25}{64} \approx 0.391$. Without: $\frac{5}{8} \cdot \frac{4}{7} = \frac{5}{14} \approx 0.357$.
88. a) Independent: $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$. b) Dependent, since the tickets are kept: $\frac{3}{15} \cdot \frac{2}{14} = \frac{1}{35}$.
89. With replacement, $\frac{9}{25} = 0.36$, beats without replacement, $\frac{3}{10} = 0.30$, by $0.06$ — six percentage points.
90. Without replacement is better now: $\frac{3}{10} + \frac{3}{10} = \frac{6}{10} = 0.60$ against $\frac{6}{25} + \frac{6}{25} = \frac{12}{25} = 0.48$. The answer flips because you now want the two draws to differ, and removing the first marble makes the other color a larger share of what remains.
91. With replacement: $\frac{500}{1000} \cdot \frac{500}{1000} = \frac{1}{4} = 0.25$. Without: $\frac{500}{1000} \cdot \frac{499}{999} \approx 0.2497$. Removing one marble from a thousand barely changes the share of red, so the second factor is almost unchanged. In a five-marble bag one marble is a fifth of everything, so removing it moves the second factor a lot.
92. With replacement: $\frac{13}{52} \cdot \frac{13}{52} = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$. Without: $\frac{13}{52} \cdot \frac{12}{51} = \frac{1}{4} \cdot \frac{4}{17} = \frac{1}{17}$. With replacement is greater, since $\frac{1}{16} > \frac{1}{17}$.
93. With replacement: $\frac{5}{25} \cdot \frac{5}{25} = \frac{1}{5} \cdot \frac{1}{5} = \frac{1}{25} = 0.04$. Without: $\frac{5}{25} \cdot \frac{4}{24} = \frac{1}{5} \cdot \frac{1}{6} = \frac{1}{30} \approx 0.033$.
94. Both rules multiply — that is what they have in common. What differs is the second factor: for independent events it is the original probability, and for dependent events it is recounted from a collection with one fewer item. Addition appears only when an event can be met in two separate ways, such as "both black or both blue."
95. Any situation in which removing the first item cannot change the second probability. For example, a bag of $5$ red marbles only: $P(\text{both red}) = 1$ whether or not the first is replaced. Another acceptable answer: if the question asks only about the first draw, replacement is irrelevant.

Exit ticket 15.4

96. With replacement $\frac{6}{10} \cdot \frac{6}{10} = \frac{9}{25} = 0.36$; without replacement $\frac{6}{10} \cdot \frac{5}{9} = \frac{1}{3} \approx 0.333$. With replacement is larger.
97. Similarity: both find $P(A \text{ and } B)$ by multiplying two probabilities, and both use the same first factor. Difference: the dependent rule uses a second factor recounted from a collection one item smaller.
98. Dependent. The second marble cannot be the first one, so the counts behave exactly as they do without replacement.
99. Ask whether the first item went back. If it did, the second denominator equals the first. If it did not, the second denominator is one less — and check whether the numerator also dropped, which happens only when the item taken was of the kind you are now counting.

Chapter 15 Review

Part A — Independent or dependent, and the role of replacement (8.PS.1a)

100. a) independent b) dependent c) independent d) independent e) dependent
101. $5$ red, $4$ green, $1$ white; total $10$. Replacement restores everything.
102. $5$ red, $3$ green, $1$ white; total $9$.
103. Replacement decides whether the second probability is computed from the original collection or from a reduced one. Without replacement the denominator drops by one, and the numerator also drops by one whenever the item removed was of the kind being counted — so both parts of the fraction can change.
104. $51$
105. Yes, independent. The dice do not interact and nothing is removed; each die still shows six equally likely faces regardless of the other.
106. Without replacement. The two marbles are different marbles, so the second one is chosen from a collection that no longer contains the first.
107. Replacement is what makes the draws independent, not dependent. Being able to draw the same marble twice is not a problem — it is precisely what keeps the second probability equal to the first. The draws are independent.

Part B — Comparing and contrasting the two cases (8.PS.1b)

108. Independent: draw 1 does not change draw 2; the item is put back; the second-stage denominator is unchanged; $P(A \text{ and } B) = P(A) \cdot P(B)$. Dependent: draw 1 does change draw 2; the item is kept out; the second-stage denominator is one smaller; $P(A \text{ and } B) = P(A) \cdot P(B \text{ after } A)$.
109. With replacement $\frac{9}{25} = 0.36$; without replacement $\frac{3}{10} = 0.30$. With replacement is greater.
110. Because nothing has been removed yet when the first event happens. Replacement is a decision about what happens after the first draw, so it cannot affect the first probability.
111. With replacement: $\frac{6}{25} + \frac{6}{25} = \frac{12}{25} = 0.48$. Without replacement: $\frac{3}{10} + \frac{3}{10} = \frac{3}{5} = 0.60$. Without replacement is greater, because removing one color leaves the other color as a larger share of what remains.
112. With replacement $\frac{3}{8} \cdot \frac{3}{8} = \frac{9}{64} \approx 0.141$; without replacement $\frac{3}{8} \cdot \frac{2}{7} = \frac{6}{56} = \frac{3}{28} \approx 0.107$. With replacement is larger.
113. Removing one item changes the counts by a smaller and smaller share as the collection grows. With $5$ marbles, one marble is $20\%$ of the total; with $1000$, it is $0.1\%$. So the second factor moves less, and the two answers converge.

Part C — The probability of two independent events (8.PS.1c)

114. $\frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12}$
115. $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$
116. $\frac{1}{5} \cdot \frac{1}{5} = \frac{1}{25}$
117. $\frac{6}{10} \cdot \frac{4}{10} = \frac{24}{100} = \frac{6}{25}$
118. $\frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16}$
119. $\frac{4}{5} \cdot \frac{4}{5} = \frac{16}{25} = 0.64 = 64\%$

Part D — The probability of two dependent events (8.PS.1d)

120. $\frac{5}{8} \cdot \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$
121. $\frac{4}{52} \cdot \frac{3}{51} = \frac{1}{13} \cdot \frac{1}{17} = \frac{1}{221}$
122. $\frac{3}{10} \cdot \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$
123. $\frac{5}{12} \cdot \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$
124. $P(\text{both black}) = \frac{6}{10} \cdot \frac{5}{9} = \frac{1}{3}$ and $P(\text{both blue}) = \frac{4}{10} \cdot \frac{3}{9} = \frac{2}{15}$, so $P(\text{match}) = \frac{5}{15} + \frac{2}{15} = \frac{7}{15}$.
125. $\frac{4}{20} \cdot \frac{3}{19} = \frac{1}{5} \cdot \frac{3}{19} = \frac{3}{95}$

Part E — Mixed application and reasoning

126. Tree 1, with replacement. First-stage branches $\frac{3}{5}$ (red) and $\frac{2}{5}$ (blue); every second-stage branch is again $\frac{3}{5}$ and $\frac{2}{5}$.

$$P(RR) = \frac{9}{25} \qquad P(RB) = \frac{6}{25} \qquad P(BR) = \frac{6}{25} \qquad P(BB) = \frac{4}{25}$$

Total: $\frac{9 + 6 + 6 + 4}{25} = \frac{25}{25} = 1$.

Tree 2, without replacement. First-stage branches are the same, $\frac{3}{5}$ and $\frac{2}{5}$. Below red: $\frac{2}{4}$ and $\frac{2}{4}$. Below blue: $\frac{3}{4}$ and $\frac{1}{4}$.

$$P(RR) = \frac{3}{10} \qquad P(RB) = \frac{3}{10} \qquad P(BR) = \frac{3}{10} \qquad P(BB) = \frac{1}{10}$$

Total: $\frac{3 + 3 + 3 + 1}{10} = \frac{10}{10} = 1$. Both totals are $1$, and only the second stage differs between the trees.

127. $\frac{4}{9} \cdot \frac{3}{8} = \frac{12}{72} = \frac{1}{6} \approx 0.167$
128. $\frac{4}{9} \cdot \frac{4}{9} = \frac{16}{81} \approx 0.198$. Larger than item 127, as expected: replacement keeps the apples' share of the bowl from shrinking, and a matching pair is always more likely with replacement.
129. Without replacement the second factor cannot use $52$ cards or $4$ aces — one ace has already been dealt, leaving $3$ aces among $51$ cards. The correct probability is $\frac{4}{52} \cdot \frac{3}{51} = \frac{1}{221}$. The student's $\frac{1}{169}$ is the answer to the with-replacement version of the question.
130. Answers vary. An acceptable response, using a bag of $4$ red and $2$ green marbles:

Independent: two draws with replacement, both red. $P = \frac{4}{6} \cdot \frac{4}{6} = \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9} \approx 0.444$.

Dependent: two draws without replacement, both red. $P = \frac{4}{6} \cdot \frac{3}{5} = \frac{12}{30} = \frac{2}{5} = 0.4$.

The independent (with-replacement) probability is larger, because the event asks for the same color twice and removing a red marble lowers red's share of what remains. Look for the same first factor in both computations, a correctly recounted second factor in the dependent case, and a comparison that names why rather than only which.