It is given that P(X ≥ 2) = 0.25 for an exponentially distributed random variable X with E[X]=1λ, where E[X] denotes the expectation of X. What is the value of λ?

(ln denotes natural logarithm)

A.

ln 2

B.

ln 4

C.

ln 3

D.

ln 0.25

Solution:

We are given:

  • X∼Exponential(λ)X \sim \text{Exponential}(\lambda)

  • P(X≥2)=0.25P(X \geq 2) = 0.25

  • E[X]=1λE[X] = \frac{1}{\lambda}

  • Find λ\lambda

Step 1: Use the exponential distribution formula

For an exponential random variable:

P(X≥x)=e−λxP(X \geq x) = e^{-\lambda x}

Given:

P(X≥2)=0.25⇒e−2λ=0.25P(X \geq 2) = 0.25 \Rightarrow e^{-2\lambda} = 0.25

Step 2: Solve for λ\lambda

Take the natural logarithm on both sides:

−2λ=ln⁡(0.25)⇒λ=−12ln⁡(0.25)-2\lambda = \ln(0.25) \Rightarrow \lambda = -\frac{1}{2} \ln(0.25)

Now, recall:

ln⁡(0.25)=ln⁡(14)=−ln⁡(4)⇒λ=−12⋅(−ln⁡(4))=12ln⁡(4)

✅ Final Answer: A

Because:

ln⁡4=ln⁡(22)=2ln⁡(2)⇒12ln⁡(4)=ln⁡(2)