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In an infection under treatment with antibiotics, a certain bacterium’s lifetime in hours is described by an exponentially distributed random variable with parameter 0.15.

a) What is the probability the bacterium won't survive for 3 hours?

b) If the bacterium survives 12 hours, what is the probability the bacterium last 1 hour longer?

For part a I know that the density of exponential distribution is $λ\cdot\exp(-λ\cdot x)$. So I should integrate $λ\cdot\exp(-\lambda\cdot x)$ with the lower bound of 0 and upper bound of 3. The calculation ended up being 0.36237.

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Sure you can integrate, and you did so correctly, but did you also know what is the cumulative distribution function for an exponential distribution?

For $X\sim\mathcal{Exp}(\lambda)$ :$$\mathsf P(0

So for $\lambda=0.15$, $~\mathsf P(0

So that's the answer to (a), which you got the hard way.

The answer to (b) is now a simple application of the definition of conditional probability.   (Although, again there is an easier method. Are you aware of the "memoryless" property and what it means?)

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    Didn't know that for part (a). So, for part (b), P(A|B) = \frac {A \bigcap B} {B}? Yes, I heard about the "memoryless" property but my memory isn't showing this up. Looking back at my notes, "The 'forgetful' or 'memory-less' nature of exponential random variables if X is an exponential random variable with parameter λ > 0: f(x) = λ⋅e^(-λ⋅x) if x ≥ 0 or 0 if x < 0." – 2017-02-28
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    How do you solve part (b) based on the "memory-less property"? @Graham Kemp – 2017-02-28
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    @Saul42 Your notes do not seem to say what is "memoryless." I see only the formula for the exponential distribution. [Here is an actual definition of "memoryless."](http://mathworld.wolfram.com/Memoryless.html) – 2017-02-28
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    @DavidK So, based on the given definition of "memory-less", P(X > s + t) = e^(-λ(s+t)). Then for problem (b), which states the bacterium has survived 12 hours and asks us for the probability of the bacterium to survive 1 hour longer should be P (X > 12 +1) = e^(-0.15(12+1))? I check back my notes and this is not correct. It should be 0.861 maybe I'm just looking at the wrong equation in WolframAlpha. – 2017-02-28
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    $P(X>12+1)$ is the unconditional probability to survive 13 hours, measured at time 0. But you have "if the bacterium survives 12 hours," so $X>12$ is a given; you want $P(X>12+1\mid X>12).$ – 2017-02-28
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    @Saul42 The definition of "memoryless" is Equation (1) on the MathWorld page. Equation (5) is just the formula of the exponential distribution, and not the conditional probability that you actually need. – 2017-02-28
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    @DavidK I believe that if I go back into the original calculation done by Graham Kemp I should get the P= 1-e^(-0.15⋅12) = 0.8347? I don't think this matches with the given answer in my problem set which is 0.861. I understand that I'm looking for the P(X > 12) can you further elaborate on this thought. – 2017-02-28
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    @Saul42 Memoryless means $P(X>s+t\mid X>t)=P(x>s).$ Let $s=1, t=12.$ Note that you want $P(x>s).$ The answer for part (a) is $P(x\leq 3).$ Do you see the difference: $>$ vs. $\leq$? For $>$ you don't subtract from $1,$ it's just $P(x>s)=\exp(-\lambda s).$ – 2017-02-28
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    Thank you for teaching this @DavidK. I appreciate your help! – 2017-02-28
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    The memoryless property is just that: the process does not remember what has gone before. The distribution of further lifetime (of an item that has not expired) is identical regardless of whenever you start measuring. $\mathsf P(X>s+t\mid X>t) = \mathsf P(X>s)$ $$\mathsf P(X>13\mid X>12) = \mathsf P(X> 1) = \exp(-0.15)$$ – 2017-02-28