Independent solution

How to solve this Exponential Distribution question

Setup

Setup

Convert the mean to an exponential rate and express the target as disjoint repeated intervals.

λ=12\lambda=\frac12
A=⋃k=0∞(k+0.75,k+1]A=\bigcup_{k=0}^{\infty}(k+0.75,k+1]

Model

Model

Write the probability of each interval by subtracting two survival values.

Pr⁡(k+0.75<T≤k+1)=e−(k+0.75)/2−e−(k+1)/2\Pr(k+0.75<T\le k+1)=e^{-(k+0.75)/2}-e^{-(k+1)/2}

Compute

Compute

Factor the common within-year term and sum the resulting geometric series.

Pr⁡(A)=∑k=0∞e−k/2(e−0.375−e−0.5)\Pr(A)=\sum_{k=0}^{\infty}e^{-k/2}\left(e^{-0.375}-e^{-0.5}\right)
Pr⁡(A)=e−0.375−e−0.51−e−0.5=0.2052476\Pr(A)=\frac{e^{-0.375}-e^{-0.5}}{1-e^{-0.5}}=0.2052476

Answer

Answer

The probability is approximately 0.205.

0.205(D)\boxed{0.205\quad\text{(D)}}