Compound Events- Independent and Dependent 1 .docx - Name I Date ... - Free Printable

Since the user did not select any specific options (ExplainAnswer, TeachTopic, or GeneratePractice are all false), I will provide only the Final Answer as per the instructions. However, to ensure accuracy, I have solved the problems internally first.

Internal Problem Solving & Verification:

1. Independent vs Dependent Events:
* 1. Rolling a die and flipping a coin: These do not affect each other. -> Independent
* 2. Drawing a card, keeping it, then drawing another: The first draw changes the total cards for the second. -> Dependent
* 3. Picking a marble, replacing it, then picking again: The total stays the same. -> Independent
* 4. Choosing a student for president, then VP from remaining: The pool shrinks. -> Dependent

2. Spinner A (4 sections: Red, Blue, Green, Yellow) and Spinner B (3 sections: 1, 2, 3):
* Total outcomes = $4 \times 3 = 12$.
* 5. P(Red and 1): Only 1 outcome (Red, 1). Probability = 1/12
* 6. P(Blue and 2): Only 1 outcome (Blue, 2). Probability = 1/12
* 7. P(Green and odd number): Odd numbers on B are 1, 3. Outcomes: (Green, 1), (Green, 3). That's 2 outcomes. Probability = 2/12 = 1/6
* 8. P(Yellow and >1): Numbers >1 on B are 2, 3. Outcomes: (Yellow, 2), (Yellow, 3). That's 2 outcomes. Probability = 2/12 = 1/6

3. Bag of Marbles (3 Red, 2 Blue, 5 Green = 10 Total):
* 9. P(Red then Blue, with replacement):
* P(Red) = 3/10
* P(Blue) = 2/10 = 1/5
* Combined: $(3/10) \times (1/5) = 3/50$ -> 3/50
* 10. P(Green then Green, with replacement):
* P(Green) = 5/10 = 1/2
* Combined: $(1/2) \times (1/2) = 1/4$ -> 1/4
* 11. P(Blue then Red, WITHOUT replacement):
* P(Blue first) = 2/10 = 1/5
* Remaining marbles = 9. Red marbles = 3.
* P(Red second) = 3/9 = 1/3
* Combined: $(1/5) \times (1/3) = 1/15$ -> 1/15
* 12. P(Green then Blue, WITHOUT replacement):
* P(Green first) = 5/10 = 1/2
* Remaining marbles = 9. Blue marbles = 2.
* P(Blue second) = 2/9
* Combined: $(1/2) \times (2/9) = 2/18 = 1/9$ -> 1/9

4. Spinner C (6 sections: 1, 2, 3, 4, 5, 6):
* 13. P(Spinning even, then odd):
* Evens: 2, 4, 6 (3 numbers). P(Even) = 3/6 = 1/2.
* Odds: 1, 3, 5 (3 numbers). P(Odd) = 3/6 = 1/2.
* Combined: $(1/2) \times (1/2) = 1/4$ -> 1/4
* 14. P(Spinning >4, then <3):
* >4: 5, 6 (2 numbers). P(>4) = 2/6 = 1/3.
* <3: 1, 2 (2 numbers). P(<3) = 2/6 = 1/3.
* Combined: $(1/3) \times (1/3) = 1/9$ -> 1/9
* 15. P(Spinning prime, then composite):
* Primes: 2, 3, 5 (3 numbers). P(Prime) = 3/6 = 1/2.
* Composites: 4, 6 (2 numbers). Note: 1 is neither. P(Composite) = 2/6 = 1/3.
* Combined: $(1/2) \times (1/3) = 1/6$ -> 1/6

5. Deck of Cards (Standard 52 cards):
* 16. P(King then Queen, WITH replacement):
* P(King) = 4/52 = 1/13.
* P(Queen) = 4/52 = 1/13.
* Combined: $(1/13) \times (1/13) = 1/169$ -> 1/169
* 17. P(Ace then Ace, WITHOUT replacement):
* P(Ace first) = 4/52 = 1/13.
* Remaining cards = 51. Remaining Aces = 3.
* P(Ace second) = 3/51 = 1/17.
* Combined: $(1/13) \times (1/17) = 1/221$ -> 1/221

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Final Answer:
1. Independent
2. Dependent
3. Independent
4. Dependent
5. 1/12
6. 1/12
7. 1/6
8. 1/6
9. 3/50
10. 1/4
11. 1/15
12. 1/9
13. 1/4
14. 1/9
15. 1/6
16. 1/169
17. 1/221