Suppose X takes the value 1 with probability P, and the value 0 with probability (1-P). The Probability distribution of X is P(X) 1-P 1 P The expected value of X, E(X) is:

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### Probability Distribution and Expected Value Calculation

**Problem Statement:**
Suppose \( X \) takes the value 1 with probability \( P \), and the value 0 with probability \( (1-P) \).

#### Probability Distribution of \( X \):

| \( X \) | \( P(X) \) |
|--------|------------|
| 0      | \( 1-P \)  |
| 1      | \( P \)    |

The expected value of \( X \), \( E(X) \), is:

#### Multiple Choice Options:
- ○ \( P \)
- ○ \( nP \)
- ○ \( P(1-P) \)
- ○ \( \frac{1-P}{P} \)

**Explanation of Calculation:**

The expected value \( E(X) \) of a discrete random variable \( X \) is given by the sum of the product of each value that \( X \) can take and its corresponding probability. Mathematically,
\[ 
E(X) = \sum (x_i \cdot P(x_i)) 
\]
For this particular case:
\[ 
E(X) = 0 \cdot (1-P) + 1 \cdot P 
\]
\[ 
E(X) = P 
\]
Thus, the correct answer is:
- ○ \( P \)

This educational example demonstrates the fundamental concept of a probability distribution and how to calculate the expected value, which is a measure of the central tendency of a random variable.

#### Graphical Representation:
There isn't a graphical representation present in this text. The probability distribution provided is in a tabular form showing the probabilities for each corresponding value of \( X \).
Transcribed Image Text:### Probability Distribution and Expected Value Calculation **Problem Statement:** Suppose \( X \) takes the value 1 with probability \( P \), and the value 0 with probability \( (1-P) \). #### Probability Distribution of \( X \): | \( X \) | \( P(X) \) | |--------|------------| | 0 | \( 1-P \) | | 1 | \( P \) | The expected value of \( X \), \( E(X) \), is: #### Multiple Choice Options: - ○ \( P \) - ○ \( nP \) - ○ \( P(1-P) \) - ○ \( \frac{1-P}{P} \) **Explanation of Calculation:** The expected value \( E(X) \) of a discrete random variable \( X \) is given by the sum of the product of each value that \( X \) can take and its corresponding probability. Mathematically, \[ E(X) = \sum (x_i \cdot P(x_i)) \] For this particular case: \[ E(X) = 0 \cdot (1-P) + 1 \cdot P \] \[ E(X) = P \] Thus, the correct answer is: - ○ \( P \) This educational example demonstrates the fundamental concept of a probability distribution and how to calculate the expected value, which is a measure of the central tendency of a random variable. #### Graphical Representation: There isn't a graphical representation present in this text. The probability distribution provided is in a tabular form showing the probabilities for each corresponding value of \( X \).
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