Activity 3: Which events are independent? Determine whether the events are independent or dependent. Then, find the probability. 1. A toy box contains 12 toys, 8 stuffed animals, and 3 board games. Maria randomly chooses 2 toys for the child she is babysiting to play with. What is the probability that she chose 2 stuffed animals as the first two choices? 2. A basket contains 6 apples, 5 bananas, 4 oranges, and 5 guavas. Dominic randomly chooses one piece of fruit, eats it, and chooses another piece of fruit. What is the probability that he chose a banana and then an apple? 3. Nick has 4 black pens, 3 blue pens, and 2 red pens in his school bag. Nick randomly picks two pens out of his school bag. What is the probability that Nick chose two blue pens, if he replaced the first pen back in his pocket before choosing a second pen?

Sure, let's analyze each of the three problems step by step to determine whether the events are independent or dependent and then calculate the required probabilities. --- ## **Problem 1** **Scenario:** - **Total Toys:** 12 - **Stuffed Animals:** 8 - **Board Games:** 3 - *(Assuming there's 1 toy of another type to make the total 12.)* - **Action:** Maria randomly chooses 2 toys **without replacement**. **Question:** What is the probability that she chooses **2 stuffed animals** as the **first two choices**? ### **Determining Independence** - **Dependent Events:** The outcome of the first choice affects the outcome of the second choice because the first toy is not replaced. - **First Choice:** The probability of selecting a stuffed animal affects the remaining number of stuffed animals and total toys for the second choice. ### **Calculating the Probability** 1. **First Choice:** - **Probability of selecting a stuffed animal:** \[ P(\text{First stuffed}) = \frac{8}{12} = \frac{2}{3} \] 2. **Second Choice (without replacement):** - **Remaining Stuffed Animals:** \(8 - 1 = 7\) - **Remaining Total Toys:** \(12 - 1 = 11\) - **Probability of selecting a stuffed animal again:** \[ P(\text{Second stuffed} \mid \text{First stuffed}) = \frac{7}{11} \] 3. **Combined Probability:** - Since the events are dependent: \[ P(\text{Both stuffed}) = P(\text{First stuffed}) \times P(\text{Second stuffed} \mid \text{First stuffed}) = \frac{2}{3} \times \frac{7}{11} = \frac{14}{33} \approx 0.4242 \text{ or } 42.42\% \] --- ## **Problem 2** **Scenario:** - **Total Fruits:** 20 - **Apples:** 6 - **Bananas:** 5 - **Oranges:** 4 - **Guavas:** 5 - **Action:** Dominic randomly chooses one piece of fruit, **eats it**, and then chooses another piece **without replacement**. **Question:** What is the probability that he chooses a **banana first** and then an **apple**? ### **Determining Independence** - **Dependent Events:** The first choice (eating a banana) affects the second choice because the total number of fruits decreases, and if a banana is eaten, it slightly affects the probabilities based on what's removed. ### **Calculating the Probability** 1. **First Choice:** - **Probability of selecting a banana:** \[ P(\text{Banana first}) = \frac{5}{20} = \frac{1}{4} \] 2. **Second Choice (after eating a banana):** - **Remaining Apples:** 6 - **Remaining Total Fruits:** \(20 - 1 = 19\) - **Probability of selecting an apple:** \[ P(\text{Apple second} \mid \text{Banana first}) = \frac{6}{19} \] 3. **Combined Probability:** - Since the events are dependent: \[ P(\text{Banana first and Apple second}) = P(\text{Banana first}) \times P(\text{Apple second} \mid \text{Banana first}) = \frac{1}{4} \times \frac{6}{19} = \frac{6}{76} = \frac{3}{38} \approx 0.0789 \text{ or } 7.89\% \] --- ## **Problem 3** **Scenario:** - **Total Pens:** 9 - **Black Pens:** 4 - **Blue Pens:** 3 - **Red Pens:** 2 - **Action:** Nick randomly picks two pens with **replacement** (i.e., he puts the first pen back before choosing the second). **Question:** What is the probability that Nick chooses **two blue pens**? ### **Determining Independence** - **Independent Events:** Replacement ensures that the first choice does not affect the second choice. The probabilities remain the same for each pick. ### **Calculating the Probability** 1. **First Choice:** - **Probability of selecting a blue pen:** \[ P(\text{First blue}) = \frac{3}{9} = \frac{1}{3} \] 2. **Second Choice (with replacement):** - **Since the pen is replaced, the probabilities remain unchanged.** - **Probability of selecting a blue pen again:** \[ P(\text{Second blue}) = \frac{3}{9} = \frac{1}{3} \] 3. **Combined Probability:** - Since the events are independent: \[ P(\text{Both blue}) = P(\text{First blue}) \times P(\text{Second blue}) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \approx 0.1111 \text{ or } 11.11\% \] --- ## **Summary** 1. **Problem 1:** - **Dependent Events** - **Probability:** \( \frac{14}{33} \) or approximately **42.42%** 2. **Problem 2:** - **Dependent Events** - **Probability:** \( \frac{3}{38} \) or approximately **7.89%** 3. **Problem 3:** - **Independent Events** - **Probability:** \( \frac{1}{9} \) or approximately **11.11%** --- If you have any further questions or need additional clarification on these problems, feel free to ask!