Independent and Dependent Events Worksheet for 7th - 8th Grade ... - Free Printable

Problem Analysis:

The task involves solving problems related to independent and dependent events. Let's break down each part of the problem step by step.

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#### Part 1: Independent Events
Independent events are those where the outcome of one event does not affect the outcome of another event. For example, rolling a die and flipping a coin are independent events.

#### Part 2: Dependent Events
Dependent events are those where the outcome of one event affects the outcome of another event. For example, drawing cards from a deck without replacement is a dependent event because the composition of the deck changes after each draw.

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Problem Details from the Image

#### Question 1: Drawing Cards from a Standard Deck
A standard deck has 52 cards: 4 suits (hearts, diamonds, clubs, spades), with 13 cards in each suit. The problem asks about probabilities of drawing specific cards under different conditions.

#### Question 2: Drawing Balls from a Bag
There are 18 black balls and 12 white balls in a bag, totaling 30 balls. The problem asks about probabilities of drawing specific sequences of balls.

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Solutions

#### Part 1: Drawing Cards from a Standard Deck

1. a. What is the probability that you choose a heart?
- There are 13 hearts in a deck of 52 cards.
- Probability = $\frac{\text{Number of hearts}}{\text{Total number of cards}} = \frac{13}{52} = \frac{1}{4}$.

2. b. What is the probability that you choose a heart, then an ace (without replacing the first card)?
- First, the probability of choosing a heart is $\frac{13}{52} = \frac{1}{4}$.
- After removing one heart, there are 51 cards left. There are 4 aces in total, so the probability of choosing an ace next is $\frac{4}{51}$.
- Since these are dependent events, we multiply the probabilities:
$$
P(\text{Heart, then Ace}) = \frac{13}{52} \times \frac{4}{51} = \frac{1}{4} \times \frac{4}{51} = \frac{1}{51}.
$$

3. c. What is the probability that you choose a heart, then an ace (with replacement)?
- With replacement, the events are independent.
- Probability of choosing a heart = $\frac{13}{52} = \frac{1}{4}$.
- Probability of choosing an ace = $\frac{4}{52} = \frac{1}{13}$.
- Multiply the probabilities:
$$
P(\text{Heart, then Ace with replacement}) = \frac{13}{52} \times \frac{4}{52} = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52}.
$$

4. d. What is the probability that you choose the king of diamonds, then the queen of hearts (without replacing the first card)?
- Probability of choosing the king of diamonds = $\frac{1}{52}$.
- After removing the king of diamonds, there are 51 cards left. The probability of choosing the queen of hearts next is $\frac{1}{51}$.
- Multiply the probabilities:
$$
P(\text{King of Diamonds, then Queen of Hearts}) = \frac{1}{52} \times \frac{1}{51} = \frac{1}{2652}.
$$

5. e. What is the probability that you choose the king of diamonds, then the queen of hearts (with replacement)?
- With replacement, the events are independent.
- Probability of choosing the king of diamonds = $\frac{1}{52}$.
- Probability of choosing the queen of hearts = $\frac{1}{52}$.
- Multiply the probabilities:
$$
P(\text{King of Diamonds, then Queen of Hearts with replacement}) = \frac{1}{52} \times \frac{1}{52} = \frac{1}{2704}.
$$

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#### Part 2: Drawing Balls from a Bag

1. a. What is the probability that he chooses a black ball, then a white ball (without replacing the first ball)?
- Probability of choosing a black ball first = $\frac{18}{30} = \frac{3}{5}$.
- After removing one black ball, there are 29 balls left, with 12 white balls remaining.
- Probability of choosing a white ball next = $\frac{12}{29}$.
- Multiply the probabilities:
$$
P(\text{Black, then White}) = \frac{18}{30} \times \frac{12}{29} = \frac{3}{5} \times \frac{12}{29} = \frac{36}{145}.
$$

2. b. What is the probability that he chooses a black ball, then a white ball (with replacement)?
- With replacement, the events are independent.
- Probability of choosing a black ball = $\frac{18}{30} = \frac{3}{5}$.
- Probability of choosing a white ball = $\frac{12}{30} = \frac{2}{5}$.
- Multiply the probabilities:
$$
P(\text{Black, then White with replacement}) = \frac{18}{30} \times \frac{12}{30} = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}.
$$

3. c. What is the probability that he chooses two black balls (without replacing the first ball)?
- Probability of choosing a black ball first = $\frac{18}{30} = \frac{3}{5}$.
- After removing one black ball, there are 29 balls left, with 17 black balls remaining.
- Probability of choosing a black ball next = $\frac{17}{29}$.
- Multiply the probabilities:
$$
P(\text{Two Black Balls}) = \frac{18}{30} \times \frac{17}{29} = \frac{3}{5} \times \frac{17}{29} = \frac{51}{145}.
$$

4. d. What is the probability that he chooses two black balls (with replacement)?
- With replacement, the events are independent.
- Probability of choosing a black ball = $\frac{18}{30} = \frac{3}{5}$.
- Multiply the probabilities:
$$
P(\text{Two Black Balls with replacement}) = \frac{18}{30} \times \frac{18}{30} = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}.
$$

5. e. What is the probability that his choices are all black?
- This is the same as part (d) since "all black" implies two black balls.
- Using the result from part (d):
$$
P(\text{All Black}) = \frac{9}{25}.
$$

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Final Answers

1. Drawing Cards from a Standard Deck:
- a. $\boxed{\frac{1}{4}}$
- b. $\boxed{\frac{1}{51}}$
- c. $\boxed{\frac{1}{52}}$
- d. $\boxed{\frac{1}{2652}}$
- e. $\boxed{\frac{1}{2704}}$

2. Drawing Balls from a Bag:
- a. $\boxed{\frac{36}{145}}$
- b. $\boxed{\frac{6}{25}}$
- c. $\boxed{\frac{51}{145}}$
- d. $\boxed{\frac{9}{25}}$
- e. $\boxed{\frac{9}{25}}$