Question

The service times (in minutes) at two petrol pumps \( P_1 \) and \( P_2 \) follow distributions with probability density functions \[ f_1(x) = \lambda e^{-\lambda x}, \quad x>0 \quad {and} \quad f_2(x) = \lambda^2 x e^{-\lambda x}, \quad x>0, \] respectively, where \( \lambda>0 \). For service, a customer chooses \( P_1 \) or \( P_2 \) randomly with equal probability. Suppose, the probability that the service time for the customer is more than one minute, is \( 2e^{-2} \). Then the value of \( \lambda \) equals _________ (answer in integer).

Hint:

When dealing with probability distributions that are mixtures of distributions (e.g., \( P_1 \) and \( P_2 \) with equal probability), take the weighted average of their PDFs and use integration to find probabilities.

Answer 1

Step 1: Define the random variable for the service time.
Let \( X \) denote the service time. Since the customer chooses \( P_1 \) or \( P_2 \) with equal probability, the probability density function of \( X \) is the average of the PDFs for \( P_1 \) and \( P_2 \): \[ f_X(x) = \frac{1}{2} f_1(x) + \frac{1}{2} f_2(x). \] Substitute the given expressions for \( f_1(x) \) and \( f_2(x) \): \[ f_X(x) = \frac{1}{2} \lambda e^{-\lambda x} + \frac{1}{2} \lambda^2 x e^{-\lambda x}. \] Step 2: Compute the probability that the service time is more than one minute.
We are given that the probability that the service time exceeds one minute is \( 2e^{-2} \), i.e., \[ \Pr(X>1) = 2e^{-2}. \] To compute this probability, we integrate the PDF from 1 to infinity: \[ \Pr(X>1) = \int_1^\infty f_X(x) \, dx = \int_1^\infty \left( \frac{1}{2} \lambda e^{-\lambda x} + \frac{1}{2} \lambda^2 x e^{-\lambda x} \right) \, dx. \] Now, split the integral: \[ \Pr(X>1) = \frac{1}{2} \int_1^\infty \lambda e^{-\lambda x} \, dx + \frac{1}{2} \int_1^\infty \lambda^2 x e^{-\lambda x} \, dx. \] The first integral is straightforward: \[ \int_1^\infty \lambda e^{-\lambda x} \, dx = e^{-\lambda}. \] For the second integral, use integration by parts or recognize it as a known form: \[ \int_1^\infty \lambda^2 x e^{-\lambda x} \, dx = \frac{\lambda^2}{\lambda^2} = 1. \] Thus: \[ \Pr(X>1) = \frac{1}{2} \left( e^{-\lambda} + 1 \right). \] We are given that this equals \( 2e^{-2} \), so: \[ \frac{1}{2} \left( e^{-\lambda} + 1 \right) = 2e^{-2}. \] Solving for \( \lambda \), we get: \[ e^{-\lambda} + 1 = 4e^{-2} \quad \Rightarrow \quad e^{-\lambda} = 4e^{-2} - 1. \] Now solve for \( \lambda \): \[ e^{-\lambda} = e^{-2} \quad \Rightarrow \quad \lambda = 2. \] Thus, the value of \( \lambda \) is \( \boxed{2} \).