Question
Scatter Plots & Data Analysis Unit Test 21 of 2321 of 23 Items Question Use the table to answer the question. Probability of Event #1 Probability of Event #2 Probability of Event #1 and Event #2 With Replacement A B 47⋅37 Without Replacement C D 47⋅36 Suppose you have seven cards—four are white and three are orange. You randomly draw two cards. Event #1: drawing a white card Event #2: drawing an orange card Compare and contrast the probability of Event #1 and Event #2 if the events are independent and if the events are dependent. Which of the following is true? (1 point) Responses The probabilities of cells B and D are the same, and the probabilities of the dependent events are higher. The probabilities of cells B and D are the same, and the probabilities of the dependent events are higher. The probabilities of cells B and D are the same, and the probabilities of the independent events are higher. The probabilities of cells B and D are the same, and the probabilities of the independent events are higher. The probabilities of cells A and C are the same, and the probabilities of the dependent events are higher. The probabilities of cells A and C are the same, and the probabilities of the dependent events are higher. The probabilities of cells A and C are the same, and the probabilities of the independent events are higher.
Answers
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GPT-4o mini
To resolve this question, we first need to understand the implications of independent and dependent events, particularly in the context of drawing cards.
### Definitions
- **Independent Events**: The occurrence of one event does not affect the occurrence of the other event. For example, if you replace a card after drawing it, the probabilities remain constant.
- **Dependent Events**: The occurrence of one event affects the occurrence of the other event. For example, if you do not replace a card after drawing it, the probabilities change after the first draw.
### Given Information
- There are 7 cards: 4 white and 3 orange.
- We are analyzing two events:
- Event #1: Drawing a white card
- Event #2: Drawing an orange card
### Probability with Replacement (Independent Events)
Let's calculate the probabilities if the cards are replaced:
- Probability of drawing a white card (Event #1):
\[ P(A) = \frac{4}{7} \]
- Probability of drawing an orange card (Event #2):
\[ P(B) = \frac{3}{7} \]
Since the events are independent, the combined probability for both events occurring:
\[ P(A \cap B) = P(A) \times P(B) = \frac{4}{7} \times \frac{3}{7} = \frac{12}{49} \]
### Probability Without Replacement (Dependent Events)
Now, let’s calculate the probabilities if the cards are not replaced:
- Probability of drawing a white card first (Event #1):
\[ P(A) = \frac{4}{7} \]
- If a white card is drawn first, there are now 6 cards left (3 white and 3 orange):
- Probability of drawing an orange card second (Event #2):
\[ P(B|A) = \frac{3}{6} = \frac{1}{2} \]
So, the probability of drawing a white card first and an orange card second:
\[ P(A \cap B) = P(A) \times P(B|A) = \frac{4}{7} \times \frac{1}{2} = \frac{4}{14} = \frac{2}{7} \]
Alternatively, if an orange card is drawn first, the calculations would be:
- Probability of drawing an orange card first (Event #2):
\[ P(B) = \frac{3}{7} \]
- Now, there are 6 cards left (4 white and 2 orange):
- Probability of drawing a white card second:
\[ P(A|B) = \frac{4}{6} = \frac{2}{3} \]
So, the probability of drawing an orange card first and a white card second:
\[ P(B \cap A) = P(B) \times P(A|B) = \frac{3}{7} \times \frac{2}{3} = \frac{6}{21} = \frac{2}{7} \]
### Summary of Findings
- The probabilities of both combinations when drawn with replacement:
- Independently (with replacement): Probabilities remain consistent.
- The probabilities without replacement affect future draws, due to the reduced number of cards.
### Response to the Multiple Choice
Based on the scenarios:
1. **Probabilities of cells B and D** are from different scenarios (independent vs. dependent) and thus are not the same.
2. **Probabilities of cells A and C** (first draw's probability) are the same in both scenarios (drawing a white card or orange card).
3. The dependent situations actually lower the probability of the second draw since the deck size is reduced.
**Final Answer:**
The probabilities of cells A and C are the same, and the probabilities of the dependent events are higher.
### Definitions
- **Independent Events**: The occurrence of one event does not affect the occurrence of the other event. For example, if you replace a card after drawing it, the probabilities remain constant.
- **Dependent Events**: The occurrence of one event affects the occurrence of the other event. For example, if you do not replace a card after drawing it, the probabilities change after the first draw.
### Given Information
- There are 7 cards: 4 white and 3 orange.
- We are analyzing two events:
- Event #1: Drawing a white card
- Event #2: Drawing an orange card
### Probability with Replacement (Independent Events)
Let's calculate the probabilities if the cards are replaced:
- Probability of drawing a white card (Event #1):
\[ P(A) = \frac{4}{7} \]
- Probability of drawing an orange card (Event #2):
\[ P(B) = \frac{3}{7} \]
Since the events are independent, the combined probability for both events occurring:
\[ P(A \cap B) = P(A) \times P(B) = \frac{4}{7} \times \frac{3}{7} = \frac{12}{49} \]
### Probability Without Replacement (Dependent Events)
Now, let’s calculate the probabilities if the cards are not replaced:
- Probability of drawing a white card first (Event #1):
\[ P(A) = \frac{4}{7} \]
- If a white card is drawn first, there are now 6 cards left (3 white and 3 orange):
- Probability of drawing an orange card second (Event #2):
\[ P(B|A) = \frac{3}{6} = \frac{1}{2} \]
So, the probability of drawing a white card first and an orange card second:
\[ P(A \cap B) = P(A) \times P(B|A) = \frac{4}{7} \times \frac{1}{2} = \frac{4}{14} = \frac{2}{7} \]
Alternatively, if an orange card is drawn first, the calculations would be:
- Probability of drawing an orange card first (Event #2):
\[ P(B) = \frac{3}{7} \]
- Now, there are 6 cards left (4 white and 2 orange):
- Probability of drawing a white card second:
\[ P(A|B) = \frac{4}{6} = \frac{2}{3} \]
So, the probability of drawing an orange card first and a white card second:
\[ P(B \cap A) = P(B) \times P(A|B) = \frac{3}{7} \times \frac{2}{3} = \frac{6}{21} = \frac{2}{7} \]
### Summary of Findings
- The probabilities of both combinations when drawn with replacement:
- Independently (with replacement): Probabilities remain consistent.
- The probabilities without replacement affect future draws, due to the reduced number of cards.
### Response to the Multiple Choice
Based on the scenarios:
1. **Probabilities of cells B and D** are from different scenarios (independent vs. dependent) and thus are not the same.
2. **Probabilities of cells A and C** (first draw's probability) are the same in both scenarios (drawing a white card or orange card).
3. The dependent situations actually lower the probability of the second draw since the deck size is reduced.
**Final Answer:**
The probabilities of cells A and C are the same, and the probabilities of the dependent events are higher.