Question

Let \(p\) be the probability of success in one trial, \(0 < p < 1\). If \(X\) is a random variable representing the number of trials until the first success occurs and \[ P(X=k)=\lambda(1-p)^{k-1},\quad k=1,2,3,\dots \] then \(\lambda = \):

• A. \(1-p\)
• B. \(p\)
• C. \(p^2\)
• D. \(1/p\)

Hint:

Geometric distributions are always normalized using the sum of infinite geometric series.

Answer 1

The Correct Option is B

Concept: The random variable \(X\) follows a geometric distribution. A valid probability mass function must satisfy: \[ \sum_{k=1}^{\infty} P(X=k) = 1 \]

Step 1: Apply normalization condition.
\[ \sum_{k=1}^{\infty} \lambda (1-p)^{k-1} = 1 \] Factor out \(\lambda\): \[ \lambda \sum_{k=1}^{\infty} (1-p)^{k-1} = 1 \]

Step 2: Evaluate geometric series.
The series is geometric with: \[ a = 1,\quad r = (1-p) \] Sum of infinite geometric series: \[ \sum_{k=1}^{\infty} (1-p)^{k-1} = \frac{1}{1-(1-p)} = \frac{1}{p} \] So: \[ \lambda \cdot \frac{1}{p} = 1 \]

Step 3: Solve for \(\lambda\).
\[ \lambda = p \] Final Answer: \[ \boxed{p} \]