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 the questions step by step, using the provided context.

---

#### Question 1: Independent Events
- Task: Determine whether the given scenarios are independent or dependent events.
- Scenarios:
1. Roll a die, then flip a coin.
2. Take a marble out of a bag, do not put it back, then take another marble.
3. Choose a person from a group of 10 people, then choose another person without putting the first one back.
4. Choose a card from a deck of cards, put it back, then choose another card.

- Solution:
- Scenario 1: Rolling a die and flipping a coin are independent events because the outcome of one does not affect the outcome of the other.
- Scenario 2: Taking a marble out of a bag and then taking another without replacing the first is a dependent event because the outcome of the first draw affects the probability of the second draw.
- Scenario 3: Choosing a person from a group and then choosing another without replacing the first is a dependent event because the first choice reduces the number of people available for the second choice.
- Scenario 4: Choosing a card from a deck, putting it back, and then choosing another card is an independent event because the card is replaced, so the probabilities remain the same for each draw.

---

#### Question 2: Dependent Events
- Task: A student chooses three cards at random from a standard deck of 52 cards without putting them back. Determine the probabilities of the following events:
1. All three cards are hearts.
2. The first card is a heart, the second is a diamond, and the third is a spade.
3. The first card is the king of diamonds, the second is the queen of clubs, and the third is the jack of hearts.

- Solution:
- Total cards in a deck: 52
- Hearts in a deck: 13
- Diamonds in a deck: 13
- Spades in a deck: 13

Part (a): All three cards are hearts
- Probability of the first card being a heart: \( \frac{13}{52} \)
- Probability of the second card being a heart (after one heart is removed): \( \frac{12}{51} \)
- Probability of the third card being a heart (after two hearts are removed): \( \frac{11}{50} \)
- Combined probability:
\[
P(\text{All three hearts}) = \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} = \frac{1716}{132600} = \frac{143}{11050}
\]

Part (b): First card is a heart, second is a diamond, third is a spade
- Probability of the first card being a heart: \( \frac{13}{52} \)
- Probability of the second card being a diamond (after one heart is removed): \( \frac{13}{51} \)
- Probability of the third card being a spade (after one heart and one diamond are removed): \( \frac{13}{50} \)
- Combined probability:
\[
P(\text{Heart, Diamond, Spade}) = \frac{13}{52} \times \frac{13}{51} \times \frac{13}{50} = \frac{2197}{132600}
\]

Part (c): First card is the king of diamonds, second is the queen of clubs, third is the jack of hearts
- Probability of the first card being the king of diamonds: \( \frac{1}{52} \)
- Probability of the second card being the queen of clubs (after one card is removed): \( \frac{1}{51} \)
- Probability of the third card being the jack of hearts (after two cards are removed): \( \frac{1}{50} \)
- Combined probability:
\[
P(\text{King of Diamonds, Queen of Clubs, Jack of Hearts}) = \frac{1}{52} \times \frac{1}{51} \times \frac{1}{50} = \frac{1}{132600}
\]

---

#### Question 3: Dependent Events with Balls
- Task: A box contains 18 black balls and 12 white balls. Every morning, the choices are as follows:
1. What is the probability that he chooses a black ball, then a white ball?
2. What is the probability that he chooses a white ball, then a black ball?
3. What is the probability that he chooses two black balls?
4. What is the probability that he chooses two white balls?
5. What is the probability that his choices are all black?

- Solution:
- Total balls: 18 black + 12 white = 30 balls

Part (a): Probability of choosing a black ball, then a white ball
- Probability of first ball being black: \( \frac{18}{30} \)
- Probability of second ball being white (after one black ball is removed): \( \frac{12}{29} \)
- Combined probability:
\[
P(\text{Black, White}) = \frac{18}{30} \times \frac{12}{29} = \frac{216}{870} = \frac{36}{145}
\]

Part (b): Probability of choosing a white ball, then a black ball
- Probability of first ball being white: \( \frac{12}{30} \)
- Probability of second ball being black (after one white ball is removed): \( \frac{18}{29} \)
- Combined probability:
\[
P(\text{White, Black}) = \frac{12}{30} \times \frac{18}{29} = \frac{216}{870} = \frac{36}{145}
\]

Part (c): Probability of choosing two black balls
- Probability of first ball being black: \( \frac{18}{30} \)
- Probability of second ball being black (after one black ball is removed): \( \frac{17}{29} \)
- Combined probability:
\[
P(\text{Black, Black}) = \frac{18}{30} \times \frac{17}{29} = \frac{306}{870} = \frac{51}{145}
\]

Part (d): Probability of choosing two white balls
- Probability of first ball being white: \( \frac{12}{30} \)
- Probability of second ball being white (after one white ball is removed): \( \frac{11}{29} \)
- Combined probability:
\[
P(\text{White, White}) = \frac{12}{30} \times \frac{11}{29} = \frac{132}{870} = \frac{22}{145}
\]

Part (e): Probability that his choices are all black
- This is the same as choosing two black balls, which we calculated in Part (c):
\[
P(\text{All Black}) = \frac{51}{145}
\]

---

Final Answers:

1. Independent/Dependent Events:
- Scenario 1: Independent
- Scenario 2: Dependent
- Scenario 3: Dependent
- Scenario 4: Independent

2. Card Probabilities:
- (a) \( \boxed{\frac{143}{11050}} \)
- (b) \( \boxed{\frac{2197}{132600}} \)
- (c) \( \boxed{\frac{1}{132600}} \)

3. Ball Probabilities:
- (a) \( \boxed{\frac{36}{145}} \)
- (b) \( \boxed{\frac{36}{145}} \)
- (c) \( \boxed{\frac{51}{145}} \)
- (d) \( \boxed{\frac{22}{145}} \)
- (e) \( \boxed{\frac{51}{145}} \)