Compound Probability quiz with multiple-choice questions involving marbles, spinners, and dice.

Let’s solve each problem one by one with clear explanations.

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Problem 1:

> A bag contains 5 blue marbles, 4 red marbles, and 3 orange marbles. Mrs. Cox picks one without looking, replaces it and picks another one. What is the probability that she picks two that are not orange?

✔ Step 1: Total marbles
5 blue + 4 red + 3 orange = 12 marbles

✔ Step 2: Marbles that are NOT orange
Blue + Red = 5 + 4 = 9 marbles

✔ Step 3: Probability of picking one marble that is not orange
= 9/12 = 3/4

✔ Step 4: Since she REPLACES the marble, the second pick is independent and has the same probability.
So, probability of picking two non-orange marbles (with replacement) is:
= (3/4) × (3/4) = 9/16

✔ Answer: C) 9/16

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Problem 2:

> You have a bag of 17 marbles. Four are blue, 6 are green, 2 are red, and the others are yellow. What is the probability of drawing a green marble, putting it aside, and then drawing another green marble?

✔ Step 1: Find number of yellow marbles
Total = 17
Blue = 4, Green = 6, Red = 2 → 4+6+2 = 12
Yellow = 17 - 12 = 5

✔ Step 2: First draw — green marble
Probability = 6/17

✔ Step 3: Put it aside → now 16 marbles left, 5 green left
Probability of second green = 5/16

✔ Step 4: Multiply for dependent events
= (6/17) × (5/16) = 30 / 272 → simplify

Divide numerator and denominator by 2:
= 15/136

✔ Answer: C) 15/136

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Problem 3:

> In the game that Tony is playing, there is a 1/8 chance the spinner will land on red and a 3/8 chance that the spinner will land on yellow. What is the probability that Tony will NOT spin red, then will spin red?

✔ Step 1: Probability of NOT spinning red
Total probability = 1
P(red) = 1/8 → P(not red) = 1 - 1/8 = 7/8

✔ Step 2: Then spin red → P(red) = 1/8

✔ Step 3: Multiply (assuming independent spins)
= (7/8) × (1/8) = 7/64

✔ Answer: B) 7/64

*(Note: The yellow probability is extra info — not needed since we’re only concerned with red vs not red.)*

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Problem 4:

> A number cube is rolled twice. What is the probability of getting a 6 on the first roll, then a number less than 5 on the second roll?

✔ Step 1: First roll — get a 6
On a standard die (1–6), P(6) = 1/6

✔ Step 2: Second roll — number less than 5
Numbers < 5: 1, 2, 3, 4 → 4 outcomes → P = 4/6 = 2/3

✔ Step 3: Multiply (independent events)
= (1/6) × (2/3) = 2/18 = 1/9

✔ Answer: B) 1/9

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Problem 5:

> Ricardo has the spinner pictured here and a bag of marbles filled with 2 red marbles, 3 green marbles, and 3 blue marbles. What is the probability that Ricardo spins red on the spinner and picks a red marble out of the bag?

⚠️ *The image isn’t fully visible, but from context and typical spinner problems, the spinner likely has equal sections — and the arrow points to “yellow” and “red” is shown as half the circle.*

Looking at the spinner diagram description:

> It shows a spinner divided into 4 equal parts? Or maybe 2? But from the image description: “red” and “yellow” are labeled, and arrow points to yellow.

But in many such problems, if red is half the spinner, then P(red) = 1/2.

Let’s assume based on common design: Spinner has 2 equal sections — red and yellow → so P(red) = 1/2

✔ Step 1: Spinner — P(red) = 1/2

✔ Step 2: Bag of marbles — total = 2 red + 3 green + 3 blue = 8 marbles
P(pick red marble) = 2/8 = 1/4

✔ Step 3: Multiply (independent events)
= (1/2) × (1/4) = 1/8

But wait — none of the options are shown for Q5 in your text. You only listed up to Q4 options.

However, since you asked to solve the problem, and assuming the spinner is half red (as visually implied), the answer is 1/8.

If the spinner had different proportions (e.g., 1/4 red), the answer would change — but without the full image, we go with most logical assumption.

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## ✔ Final Answers:

1. C) 9/16
2. C) 15/136
3. B) 7/64
4. B) 1/9
5. Assuming spinner is half red → 1/8 *(but no options provided)*

Let me know if you can share the spinner image details for Q5 — I’ll adjust accordingly!