Let `E` be an event which is neither a certainty nor an impossibility. If probability is such that `P(E)=1+lambda+lambda^2` and `P(E^(prime))=(1+lambda)^2` in terms of an unknown `lambdadot` Then `P(E)` is equal to

Options

A. 1
B. `(3)/(4)`
C. `(1)/(4)`
D. none of these

Explanation

To solve the problem step by step, we will analyze the given probabilities and use the properties of probability.

### Step 1: Understand the given probabilities
We are given:
- \( P(E) = 1 + \lambda + \lambda^2 \)
- \( P(E') = (1 + \lambda)^2 \)

### Step 2: Use the property of probabilities
We know that the sum of the probability of an event and its complement is equal to 1:
\[ P(E) + P(E') = 1 \]

### Step 3: Substitute the given probabilities into the equation
Substituting the expressions for \( P(E) \) and \( P(E') \):
\[ (1 + \lambda + \lambda^2) + (1 + \lambda)^2 = 1 \]

### Step 4: Expand \( P(E') \)
Now, expand \( P(E') \):
\[ (1 + \lambda)^2 = 1 + 2\lambda + \lambda^2 \]
So the equation becomes:
\[ 1 + \lambda + \lambda^2 + 1 + 2\lambda + \lambda^2 = 1 \]

### Step 5: Combine like terms
Combine the terms:
\[ 2 + 3\lambda + 2\lambda^2 = 1 \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ 2\lambda^2 + 3\lambda + 2 - 1 = 0 \]
\[ 2\lambda^2 + 3\lambda + 1 = 0 \]

### Step 7: Factor the quadratic equation
We can factor the quadratic equation:
\[ (2\lambda + 1)(\lambda + 1) = 0 \]

### Step 8: Solve for \( \lambda \)
Setting each factor to zero gives:
1. \( 2\lambda + 1 = 0 \) → \( \lambda = -\frac{1}{2} \)
2. \( \lambda + 1 = 0 \) → \( \lambda = -1 \)

### Step 9: Substitute values of \( \lambda \) back into \( P(E) \)
Now we will substitute these values back into \( P(E) \):
1. For \( \lambda = -\frac{1}{2} \):
\[ P(E) = 1 - \frac{1}{2} + \left(-\frac{1}{2}\right)^2 = 1 - \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \]

2. For \( \lambda = -1 \):
\[ P(E) = 1 - 1 + (-1)^2 = 1 - 1 + 1 = 1 \]

### Step 10: Determine the valid probability
Since \( E \) is neither a certainty nor an impossibility, we reject \( P(E) = 1 \). Therefore, the valid probability is:
\[ P(E) = \frac{3}{4} \]

Thus, the final answer is:
\[ \boxed{\frac{3}{4}} \]