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A bag contains four tickets marked with numbers `112,121,211,` and `222`. One ticket is drawn at random from the bag. Let `E_(i)(i=1,2,3)` denote the event that `ith` digit on the ticket is `2`. Then

A

`E_(1)` and `E_(2)` are independent

B

`E_(2)` and `E_(3)` are independent

C

`E_(3)` and `E_(1)` are independent

D

`E_(1)`,`E_(2)`.`E_(3)` are independent

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the events \( E_1, E_2, \) and \( E_3 \) based on the tickets drawn from the bag. ### Step 1: Define the Events Let’s define the events based on the digits of the tickets: - \( E_1 \): The event that the first digit of the ticket is 2. - \( E_2 \): The event that the second digit of the ticket is 2. - \( E_3 \): The event that the third digit of the ticket is 2. ### Step 2: List the Tickets The tickets in the bag are: 1. 112 2. 121 3. 211 4. 222 ### Step 3: Determine the Outcomes for Each Event Now we will determine which tickets correspond to each event: - **For \( E_1 \)** (First digit is 2): - Tickets: None (0 tickets) - **For \( E_2 \)** (Second digit is 2): - Tickets: 112, 222 (2 tickets) - **For \( E_3 \)** (Third digit is 2): - Tickets: 112, 121, 222 (3 tickets) ### Step 4: Calculate Probabilities Now we calculate the probabilities of each event: - \( P(E_1) = \frac{0}{4} = 0 \) - \( P(E_2) = \frac{2}{4} = \frac{1}{2} \) - \( P(E_3) = \frac{3}{4} \) ### Step 5: Check Independence of Events Two events \( E_i \) and \( E_j \) are independent if: \[ P(E_i \cap E_j) = P(E_i) \cdot P(E_j) \] 1. **Check \( E_1 \) and \( E_2 \)**: - \( P(E_1 \cap E_2) = 0 \) - \( P(E_1) \cdot P(E_2) = 0 \cdot \frac{1}{2} = 0 \) - **Independent**. 2. **Check \( E_2 \) and \( E_3 \)**: - \( P(E_2 \cap E_3) = P(\{112, 222\} \cap \{112, 121, 222\}) = P(\{112, 222\}) = \frac{2}{4} = \frac{1}{2} \) - \( P(E_2) \cdot P(E_3) = \frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8} \) - **Not Independent**. 3. **Check \( E_1 \) and \( E_3 \)**: - \( P(E_1 \cap E_3) = 0 \) - \( P(E_1) \cdot P(E_3) = 0 \cdot \frac{3}{4} = 0 \) - **Independent**. ### Step 6: Conclusion From the checks above, we conclude: - \( E_1 \) and \( E_2 \) are independent. - \( E_2 \) and \( E_3 \) are not independent. - \( E_1 \) and \( E_3 \) are independent. Therefore, the correct options are that \( E_1 \) and \( E_2 \) are independent, \( E_1 \) and \( E_3 \) are independent, but \( E_2 \) and \( E_3 \) are not independent.

To solve the problem, we need to analyze the events \( E_1, E_2, \) and \( E_3 \) based on the tickets drawn from the bag. ### Step 1: Define the Events Let’s define the events based on the digits of the tickets: - \( E_1 \): The event that the first digit of the ticket is 2. - \( E_2 \): The event that the second digit of the ticket is 2. - \( E_3 \): The event that the third digit of the ticket is 2. ...
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