The pdf for T∼Exponential(λ)T\sim \text{Exponential}(\lambda) is

f(t∣λ)=λe−λt⋅It>0f(t|\lambda)=\lambda e^{-\lambda t} \cdot \mathbb{I}_{t\gt 0}

Its cdf is

F(t∣λ)=(1−e−λt)⋅It>0F(t|\lambda)=(1-e^{-\lambda t})\cdot \mathbb{I}_{t\gt 0}

which implies the survivor function is

S(t):=P(T>t)=1−F(t∣λ)=e−λtS(t):=\mathbb{P}(T\gt t)=1-F(t|\lambda)=e^{-\lambda t}

Example 1

A very large logging operation has serious problems keeping their skidders operating properly. The equipment fails at the rate of 33 breakdowns every 4848 hours. Assume that xx is time between breakdowns and is exponentially distributed. What is the probability of a single breakdown within 2424 hours?

So the statement gives x∼Exponential(348)x\sim \text{Exponential}(\frac{3}{48}), whose cdf F(t∣λ)=1−e−λtF(t|\lambda)=1-e^{-\lambda t}. The answer is then F(24)=1−exp⁡(−1.5)=0.7769F(24)=1-\exp(-1.5)=0.7769