Memorylessness of the Exponential Distribution

Please help me solve the following question with two parts.

$T$ is the time required to repair a machine. We have that $T$ is exponentially distributed with a mean of $\frac{1}{2}$ hours.

For the first part of the question, I am asked to find the probability that repair time exceeds $\frac{1}{2}$ hours. I find
$$P(T> \frac{1}{2})= \frac{1}{e}$$

I am a bit stuck on the second part to the question:
“What is the probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours”?

I was thinking the answer is

$$P(T\geq 12.5 \, | \, T>12)=P(T\geq .5) = \frac{1}{e} \, \text{ ,}$$

since exponential distribution is memoryless. Is this correct?

Solutions Collecting From Web of "Memorylessness of the Exponential Distribution"

Yes, this is correct, this is indeed what is meant by memorylessness.

In general, for all $s\geq 0$, $t \geq 0$, we have

$$P(T>s+t\mid T>s) = P(T>t) = e^{-\lambda t}.$$

We can prove the memorylessness by transforming the expression on the LHS using the definition of conditional probability, as follows

P(X>s+t\mid X>s) &= \frac{P\big(X>s+t\cap X>s\big)}{P(X>s)} = \frac{P(X>s+t)}{P(X>s)}
&= \frac{\int_{s+t}^{\infty}\lambda e^{-\lambda x}dx}{\int_{s}^{\infty}\lambda e^{-\lambda x}dx}= \frac{0- -e^{-\lambda (s+t)}}{0- -e^{-\lambda s}} = e^{-\lambda t} = P(X>t)

Accreditation: This is a slight modification of this answer by Tianyu Zheng