Consider a transmission system where the input, X, equals 1, 2, 3 or 4, with respective probabilities 0.1,

0.2, 0.4 and 0.3. The transmission system is imperfect. Given X = x, the output, Y, equals x or x +1, with

respective probabilities 0.7 and 0.3, when x is even. On the other hand, given X = x, the output, Y, equals

x or x –1, with respective probabilities 0.8 and 0.2, when x is odd. The system also has a feature that

reports an outcome, Z, which equals 0, 1 or 2, according as Y = X, Y > X or Y < X, respectively.

1. The value of E(Z) is nearest to

A) 0.33 B) 0.34 C) 0.35 D) 0.36 E) 0.37


explain very simply, step by step, im very stupid

Let's break down the problem step by step. We need to find the expected value \( E(Z) \), where \( Z \) is a variable that takes values 0, 1, or 2 depending on the relationship between \( Y \) and \( X \):

- \( Z = 0 \) if \( Y = X \)
- \( Z = 1 \) if \( Y > X \)
- \( Z = 2 \) if \( Y < X \)

### Step 1: Understand the probabilities for \( X \)
The input \( X \) can be 1, 2, 3, or 4 with these probabilities:
- \( P(X = 1) = 0.1 \)
- \( P(X = 2) = 0.2 \)
- \( P(X = 3) = 0.4 \)
- \( P(X = 4) = 0.3 \)

### Step 2: Determine the behavior of \( Y \)
The output \( Y \) depends on whether \( X \) is even or odd:
- If \( X \) is even (2 or 4):
  - \( Y = X \) with probability 0.7
  - \( Y = X + 1 \) with probability 0.3
- If \( X \) is odd (1 or 3):
  - \( Y = X \) with probability 0.8
  - \( Y = X - 1 \) with probability 0.2

### Step 3: Calculate the probabilities for \( Z = 0, 1, 2 \)
We calculate \( P(Z = 0) \), \( P(Z = 1) \), and \( P(Z = 2) \) for each \( X \):

1. **For \( X = 1 \) (odd):**
   - \( Y = 1 \) (same as \( X \)): Probability 0.8 → \( Z = 0 \)
   - \( Y = 0 \) (less than \( X \)): Probability 0.2 → \( Z = 2 \)

2. **For \( X = 2 \) (even):**
   - \( Y = 2 \) (same as \( X \)): Probability 0.7 → \( Z = 0 \)
   - \( Y = 3 \) (greater than \( X \)): Probability 0.3 → \( Z = 1 \)

3. **For \( X = 3 \) (odd):**
   - \( Y = 3 \) (same as \( X \)): Probability 0.8 → \( Z = 0 \)
   - \( Y = 2 \) (less than \( X \)): Probability 0.2 → \( Z = 2 \)

4. **For \( X = 4 \) (even):**
   - \( Y = 4 \) (same as \( X \)): Probability 0.7 → \( Z = 0 \)
   - \( Y = 5 \) (greater than \( X \)): Probability 0.3 → \( Z = 1 \)

### Step 4: Combine probabilities for \( Z = 0, 1, 2 \)
Let's calculate the overall probabilities:

- \( P(Z = 0) = (0.1 \times 0.8) + (0.2 \times 0.7) + (0.4 \times 0.8) + (0.3 \times 0.7) \)
- \( P(Z = 1) = (0.2 \times 0.3) + (0.3 \times 0.3) \)
- \( P(Z = 2) = (0.1 \times 0.2) + (0.4 \times 0.2) \)

Simplifying:

- \( P(Z = 0) = 0.08 + 0.14 + 0.32 + 0.21 = 0.75 \)
- \( P(Z = 1) = 0.06 + 0.09 = 0.15 \)
- \( P(Z = 2) = 0.02 + 0.08 = 0.10 \)

### Step 5: Calculate \( E(Z) \)
The expected value \( E(Z) \) is:

\[
E(Z) = 0 \times P(Z = 0) + 1 \times P(Z = 1) + 2 \times P(Z = 2)
\]

Substitute the probabilities:

\[
E(Z) = 0 \times 0.75 + 1 \times 0.15 + 2 \times 0.10 = 0 + 0.15 + 0.20 = 0.35
\]

So, the expected value \( E(Z) \) is closest to **0.35**.

### Answer:
The value of \( E(Z) \) is nearest to **C) 0.35**.

---

Would you like more details or have any questions? Here are some follow-up questions:

1. How do we calculate the variance of \( Z \) based on the given probabilities?
2. What would happen if the probabilities for \( X \) were different?
3. Can you explain the concept of expected value in more detail?
4. How do conditional probabilities apply to this problem?
5. How would the solution change if the transmission system were perfect?

**Tip:** Always double-check your calculations, especially in problems with multiple steps involving probabilities.

Consider a transmission system where the input, X, equals 1, 2, 3 or 4, with respective probabilities 0.1, 0.2, 0.4 and 0.3. The transmission system is imperfect. Given X = x, the output, Y, equals x or x +1, with respective probabilities 0.7 and 0.3, when x is even. On the other hand, given X = x, the output, Y, equals x or x –1, with respective probabilities 0.8 and 0.2, when x is odd. The system also has a feature that reports an outcome, Z, which equals 0, 1 or 2, according as Y = X, Y > X or Y < X, respectively. The value of E(Z) is nearest to A) 0.33 B) 0.34 C) 0.35 D) 0.36 E) 0.37

Learn how to calculate the expected value (E(Z)) for a transmission system where the output is based on conditional probabilities of the input. This problem covers key concepts like probability, expected value, and conditional probability, suitable for students in Grades 11-12.

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