10-5 Independent and Dependent Events | PDF | Probability ... - Free Printable

Let’s solve each problem step by step.

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Part 1: Determine if events are dependent or independent

> Independent events: The outcome of one does NOT affect the other.
> Dependent events: The outcome of one DOES affect the other.

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1. Choosing a tie and shirt from the closet

- If you pick a tie first, then pick a shirt — your choice of tie doesn’t change what shirts are available (unless they’re matched sets, but we assume random selection).
- So, these are independent.

✔ Answer: Independent

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2. Choosing a month and tossing a coin

- Picking a month has nothing to do with flipping a coin.
- One doesn’t affect the other.
- So, these are independent.

✔ Answer: Independent

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3. Rolling two fair number cubes once, then rolling them again if you received the same number on both cubes on the first roll

- Whether you roll again depends on what happened in the first roll.
- So, the second action is *conditional* on the first → dependent.

✔ Answer: Dependent

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Part 2: Experiment = rolling a fair number cube AND tossing a fair coin

A fair number cube has sides: 1, 2, 3, 4, 5, 6 → each with probability 1/6
A fair coin has heads and tails → each with probability 1/2
Since rolling and tossing are independent, multiply probabilities.

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4. Probability of getting a 5 on the cube AND tails on the dime

P(5) = 1/6
P(tails) = 1/2
→ Multiply: (1/6) × (1/2) = 1/12

✔ Answer: 1/12

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5. Probability of even number on cube AND heads on dime

Even numbers on cube: 2, 4, 6 → 3 out of 6 → P(even) = 3/6 = 1/2
P(heads) = 1/2
→ Multiply: (1/2) × (1/2) = 1/4

✔ Answer: 1/4

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6. Probability of getting a 2 or 3 on cube AND heads on dime

P(2 or 3) = 2/6 = 1/3
P(heads) = 1/2
→ Multiply: (1/3) × (1/2) = 1/6

✔ Answer: 1/6

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Part 3: Box with marbles — 3 red, 6 blue, 1 white → total = 10 marbles

Marbles are selected one at a time, without replacement → so events are dependent. We must adjust probabilities after each draw.

We’ll calculate each probability as a product of fractions, updating the total and counts after each draw.

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7. P(blue and red)

This means: first blue, then red? Or any order? In probability like this, unless specified, “and” usually means in sequence — but since it doesn’t specify order, we should consider both orders? Wait — looking at later problems like #10 (“red and white and blue”), it seems they mean in that exact order.

But let’s check context. Problem 7 says “P(blue and red)” — likely meaning first blue, then red. But actually, in many textbooks, when they say “P(A and B)” for sequential draws without replacement, they often mean in that order unless stated otherwise. However, sometimes they mean either order. Let’s look at problem 8: “P(white and blue)” — again, probably specific order.

Wait — problem 9 is “P(red and white)” — again, likely order matters.

Actually, looking at problem 10: “P(red and white and blue)” — three in a row, so order matters.

So I think for all these, the order written is the order drawn.

So:

7. P(blue and red) → first blue, then red

Total marbles: 10
P(first blue) = 6/10
After removing one blue, left: 9 marbles (3 red, 5 blue, 1 white)
P(second red) = 3/9
→ Multiply: (6/10) × (3/9) = (6×3)/(10×9) = 18/90 = 1/5

✔ Answer: 1/5

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8. P(white and blue) → first white, then blue

P(first white) = 1/10
After removing white: 9 marbles left (3 red, 6 blue)
P(second blue) = 6/9
→ Multiply: (1/10) × (6/9) = 6/90 = 1/15

✔ Answer: 1/15

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9. P(red and white) → first red, then white

P(first red) = 3/10
After removing red: 9 marbles left (2 red, 6 blue, 1 white)
P(second white) = 1/9
→ Multiply: (3/10) × (1/9) = 3/90 = 1/30

✔ Answer: 1/30

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10. P(red and white and blue) → first red, then white, then blue

Start: 10 marbles
P(red) = 3/10
After red removed: 9 marbles (2R, 6B, 1W)
P(white) = 1/9
After white removed: 8 marbles (2R, 6B)
P(blue) = 6/8
→ Multiply: (3/10) × (1/9) × (6/8) = (3×1×6)/(10×9×8) = 18 / 720 = simplify: divide numerator and denominator by 18 → 1/40

Wait: 18 ÷ 18 = 1, 720 ÷ 18 = 40 → yes, 1/40

✔ Answer: 1/40

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11. P(red and red and blue) → first red, second red, third blue

Start: 10 marbles
P(first red) = 3/10
After first red: 9 marbles (2R, 6B, 1W)
P(second red) = 2/9
After second red: 8 marbles (1R, 6B, 1W)
P(third blue) = 6/8
→ Multiply: (3/10) × (2/9) × (6/8) = (3×2×6)/(10×9×8) = 36 / 720 = simplify: divide by 36 → 1/20

Wait: 36 ÷ 36 = 1, 720 ÷ 36 = 20 → yes, 1/20

✔ Answer: 1/20

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12. P(red and blue and blue) → first red, then blue, then blue

Start: 10
P(red) = 3/10
After red: 9 marbles (2R, 6B, 1W)
P(blue) = 6/9
After blue: 8 marbles (2R, 5B, 1W)
P(blue again) = 5/8
→ Multiply: (3/10) × (6/9) × (5/8) = (3×6×5)/(10×9×8) = 90 / 720 = 1/8

Wait: 90 ÷ 90 = 1, 720 ÷ 90 = 8 → yes, 1/8

✔ Answer: 1/8

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13. P(red and red and red) → three reds in a row

Start: 10
P(first red) = 3/10
After first red: 9 marbles (2R, ...) → P(second red) = 2/9
After second red: 8 marbles (1R, ...) → P(third red) = 1/8
→ Multiply: (3/10) × (2/9) × (1/8) = 6 / 720 = 1/120

✔ Answer: 1/120

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14. P(white and blue and blue) → first white, then blue, then blue

Start: 10
P(white) = 1/10
After white: 9 marbles (3R, 6B)
P(blue) = 6/9
After blue: 8 marbles (3R, 5B)
P(blue again) = 5/8
→ Multiply: (1/10) × (6/9) × (5/8) = (1×6×5)/(10×9×8) = 30 / 720 = 1/24

✔ Answer: 1/24

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15. P(white and red and white) → first white, then red, then white

Wait — there’s only 1 white marble! So after drawing white first, there are no whites left.

So:

P(first white) = 1/10
After white removed: 9 marbles (3R, 6B, 0W)
P(second red) = 3/9
After red removed: 8 marbles (2R, 6B, 0W)
P(third white) = 0/8 = 0
→ Whole thing = 0

✔ Answer: 0

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Final Answers:

1. Independent
2. Independent
3. Dependent
4. 1/12
5. 1/4
6. 1/6
7. 1/5
8. 1/15
9. 1/30
10. 1/40
11. 1/20
12. 1/8
13. 1/120
14. 1/24
15. 0