Question

If \( X \) is a Poisson random variable with mean \( \mu = 1 \), then the conditional probability of the event \( \{ X \geq 2 \} \) given that the event \( \{ X \geq 4 \} \) has occurred, is ……… (rounded off to two decimal places).
 

Hint:

For conditional probabilities involving subsets (\( A \subseteq B \)), the probability \( P(A \mid B) \) simplifies to 1 because \( A \cap B = B \) and \( P(A \cap B) = P(B) \).

Answer 1

Step 1: Recall the probability mass function (PMF) of a Poisson random variable. 
The PMF of a Poisson random variable \( X \) with mean \( \mu \) is given by: \[ P(X = k) = \frac{e^{-\mu} \mu^k}{k!}, \quad k = 0, 1, 2, \ldots \] Here, \( \mu = 1 \). 

Step 2: Define the conditional probability. 
The conditional probability is defined as: \[ P(X \geq 2 \mid X \geq 4) = \frac{P(X \geq 2 \cap X \geq 4)}{P(X \geq 4)}. \] Since \( X \geq 4 \) implies \( X \geq 2 \), the numerator simplifies to \( P(X \geq 4) \). Thus: \[ P(X \geq 2 \mid X \geq 4) = \frac{P(X \geq 4)}{P(X \geq 4)} = 1. \] 

Step 3: Conclusion. 
The conditional probability \( P(X \geq 2 \mid X \geq 4) \) is: \[ P(X \geq 2 \mid X \geq 4) = 1.00. \] 

Conclusion: The conditional probability is \( 1.00 \).