Independent solution

How to solve this Exponential Distribution question

Setup

Setup

Let T be the waiting time and convert the stated mean to the exponential rate.

T∼Exponential⁡(λ),λ=14T\sim\operatorname{Exponential}(\lambda),\qquad \lambda=\frac{1}{4}

Model

Model

Conditioning on no event before year 2 leaves an independent exponential residual wait with the same rate.

(T−2∣T>2)∼Exponential⁡ ⁣(14)(T-2\mid T>2)\sim\operatorname{Exponential}\!\left(\frac14\right)
Pr⁡(5<T<8∣T>2)=Pr⁡(3<T−2<6∣T>2)\Pr(5<T<8\mid T>2)=\Pr(3<T-2<6\mid T>2)

Compute

Compute

Subtract exponential survival probabilities at the two residual-time boundaries.

Pr⁡(5<T<8∣T>2)=e−3/4−e−6/4\Pr(5<T<8\mid T>2)=e^{-3/4}-e^{-6/4}
=0.4723665527−0.2231301601=0.2492363926=0.4723665527-0.2231301601=0.2492363926

Answer

Answer

The conditional probability rounds to 0.25.

0.25(C)\boxed{0.25\quad\text{(C)}}