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An urn contains four balls bearing numbe...

An urn contains four balls bearing numbers 1,2,3 and 123 respectively . A ball is drawn at random from the urn. Let `E_(p) i = 1,2,3` donote the event that digit i appears on the ball drawn
statement 1 : `P(E_(1)capE_(2)) = P(E_(1) cap E_(3)) = P(E_(2) cap E_(3)) = (1)/(4)`
Statement 2 : `P_(E_(1)) = P(E_(2)) = P(E_(3)) = (1)/(2)`

A

Statement 1 and statement 2 are both false

B

Statement 1 and statement 2 are both true

C

Statement 1 is true and statement 2 is false

D

Statement 1 is false and statement 2 is true

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the events defined in the question and calculate the probabilities accordingly. ### Step-by-Step Solution: 1. **Identify the Balls in the Urn:** The urn contains four balls with the following numbers: 1, 2, 3, and 123. 2. **Define the Events:** - Let \( E_1 \) be the event that digit 1 appears on the ball drawn. - Let \( E_2 \) be the event that digit 2 appears on the ball drawn. - Let \( E_3 \) be the event that digit 3 appears on the ball drawn. 3. **Calculate the Probability of Each Event:** - For \( E_1 \): The balls that contain the digit 1 are: 1 and 123. Therefore, the number of favorable outcomes for \( E_1 \) is 2. \[ P(E_1) = \frac{\text{Number of favorable outcomes for } E_1}{\text{Total number of balls}} = \frac{2}{4} = \frac{1}{2} \] - For \( E_2 \): The balls that contain the digit 2 are: 2 and 123. Therefore, the number of favorable outcomes for \( E_2 \) is 2. \[ P(E_2) = \frac{2}{4} = \frac{1}{2} \] - For \( E_3 \): The balls that contain the digit 3 are: 3 and 123. Therefore, the number of favorable outcomes for \( E_3 \) is 2. \[ P(E_3) = \frac{2}{4} = \frac{1}{2} \] 4. **Check Statement 2:** According to Statement 2, \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{2} \). Since we have calculated all three probabilities to be \( \frac{1}{2} \), Statement 2 is **true**. 5. **Calculate the Intersection of Events:** - For \( P(E_1 \cap E_2) \): This is the probability that both digits 1 and 2 appear on the drawn ball. The only ball that contains both digits is 123. \[ P(E_1 \cap E_2) = \frac{1}{4} \] - For \( P(E_1 \cap E_3) \): This is the probability that both digits 1 and 3 appear on the drawn ball. The only ball that contains both digits is 123. \[ P(E_1 \cap E_3) = \frac{1}{4} \] - For \( P(E_2 \cap E_3) \): This is the probability that both digits 2 and 3 appear on the drawn ball. The only ball that contains both digits is 123. \[ P(E_2 \cap E_3) = \frac{1}{4} \] 6. **Check Statement 1:** According to Statement 1, \( P(E_1 \cap E_2) = P(E_1 \cap E_3) = P(E_2 \cap E_3) = \frac{1}{4} \). Since we have calculated all three intersections to be \( \frac{1}{4} \), Statement 1 is **true**. ### Conclusion: Both Statement 1 and Statement 2 are true.
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