Multiplication Rule of Probability - Independent Practice Worksheet featuring ten probability problems involving cards, bottles, stones, shirts, blocks, lipsticks, balls, beads, shoes, and caps.

Let's solve each problem step by step using the Multiplication Rule of Probability for independent or dependent events, depending on whether the draws are with or without replacement.

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Problem 1:

Holly is going to draw two cards from a standard deck without replacement. What is the probability that the first card is a king and the second card is an ace?

- A standard deck has 52 cards.
- There are 4 kings and 4 aces.
- Drawing without replacement, so events are dependent.

Step 1: P(first card is a king) = $ \frac{4}{52} $

Step 2: After removing one king, 51 cards remain.
P(second card is an ace) = $ \frac{4}{51} $

Multiply:
$$
P(\text{king then ace}) = \frac{4}{52} \times \frac{4}{51} = \frac{1}{13} \times \frac{4}{51} = \frac{4}{663}
$$

✔ Answer: $ \boxed{\frac{4}{663}} $

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Problem 2:

Thomas has a box with 4 black bottles and 8 gray bottles. Two bottles are drawn without replacement. What is the probability that both are gray?

Total bottles = 4 + 8 = 12

P(first gray) = $ \frac{8}{12} $

After removing one gray, 7 gray left, 11 total.

P(second gray) = $ \frac{7}{11} $

$$
P(\text{both gray}) = \frac{8}{12} \times \frac{7}{11} = \frac{2}{3} \times \frac{7}{11} = \frac{14}{33}
$$

✔ Answer: $ \boxed{\frac{14}{33}} $

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Problem 3:

A jar has 4 pink, 9 orange, 5 green stones. Ryan picks one stone, records color, puts it back, then draws again. What is the probability of orange followed by green?

This is with replacement, so events are independent.

Total stones = 4 + 9 + 5 = 18

P(orange) = $ \frac{9}{18} = \frac{1}{2} $

P(green) = $ \frac{5}{18} $

$$
P(\text{orange then green}) = \frac{1}{2} \times \frac{5}{18} = \frac{5}{36}
$$

✔ Answer: $ \boxed{\frac{5}{36}} $

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Problem 4:

Henry has 3 black shirts and 7 blue shirts. Two shirts drawn without replacement. What is the probability both are black?

Total = 10 shirts

P(first black) = $ \frac{3}{10} $

P(second black) = $ \frac{2}{9} $ (since one black removed)

$$
P(\text{both black}) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}
$$

✔ Answer: $ \boxed{\frac{1}{15}} $

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Problem 5:

Andrew has 4 pink, 5 yellow, 6 green blocks. He picks one, records, puts back, then picks another. What is the probability of yellow followed by pink?

With replacement → independent.

Total = 4 + 5 + 6 = 15

P(yellow) = $ \frac{5}{15} = \frac{1}{3} $

P(pink) = $ \frac{4}{15} $

$$
P(\text{yellow then pink}) = \frac{1}{3} \times \frac{4}{15} = \frac{4}{45}
$$

✔ Answer: $ \boxed{\frac{4}{45}} $

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Problem 6:

Anna has 2 purple, 3 red, 3 pink lipsticks. She picks one, records, puts back, then picks another. What is the probability of purple followed by red?

With replacement → independent.

Total = 2 + 3 + 3 = 8

P(purple) = $ \frac{2}{8} = \frac{1}{4} $

P(red) = $ \frac{3}{8} $

$$
P(\text{purple then red}) = \frac{1}{4} \times \frac{3}{8} = \frac{3}{32}
$$

✔ Answer: $ \boxed{\frac{3}{32}} $

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Problem 7:

Jordan has 9 blue, 6 red, 5 brown balls. Two drawn without replacement. What is the probability both are brown?

Total = 9 + 6 + 5 = 20

P(first brown) = $ \frac{5}{20} = \frac{1}{4} $

P(second brown) = $ \frac{4}{19} $

$$
P(\text{both brown}) = \frac{1}{4} \times \frac{4}{19} = \frac{4}{76} = \frac{1}{19}
$$

✔ Answer: $ \boxed{\frac{1}{19}} $

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Problem 8:

Bella has 7 green, 10 yellow, 3 white beads. Two drawn without replacement. What is the probability both are yellow?

Total = 7 + 10 + 3 = 20

P(first yellow) = $ \frac{10}{20} = \frac{1}{2} $

P(second yellow) = $ \frac{9}{19} $

$$
P(\text{both yellow}) = \frac{1}{2} \times \frac{9}{19} = \frac{9}{38}
$$

✔ Answer: $ \boxed{\frac{9}{38}} $

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Problem 9:

Finn has 4 pairs of black shoes and 4 pairs of brown shoes. He picks one pair, records, puts back, then picks another. What is the probability of black followed by brown?

With replacement → independent.

Total pairs = 4 + 4 = 8

P(black) = $ \frac{4}{8} = \frac{1}{2} $

P(brown) = $ \frac{4}{8} = \frac{1}{2} $

$$
P(\text{black then brown}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
$$

✔ Answer: $ \boxed{\frac{1}{4}} $

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Problem 10:

Caleb has 7 black caps, 4 yellow caps, 9 blue caps. Two caps drawn without replacement. What is the probability both are blue?

Total = 7 + 4 + 9 = 20

P(first blue) = $ \frac{9}{20} $

P(second blue) = $ \frac{8}{19} $

$$
P(\text{both blue}) = \frac{9}{20} \times \frac{8}{19} = \frac{72}{380} = \frac{18}{95}
$$

✔ Answer: $ \boxed{\frac{18}{95}} $

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ \frac{4}{663} $ |
| 2 | $ \frac{14}{33} $ |
| 3 | $ \frac{5}{36} $ |
| 4 | $ \frac{1}{15} $ |
| 5 | $ \frac{4}{45} $ |
| 6 | $ \frac{3}{32} $ |
| 7 | $ \frac{1}{19} $ |
| 8 | $ \frac{9}{38} $ |
| 9 | $ \frac{1}{4} $ |
| 10 | $ \frac{18}{95} $ |

Let me know if you'd like these explained in decimal or percentage form!