What is the probability that the next call come in $3$ minutes provided that in the past $2$ minutes there have been no calls?

Question

At a certain customer service line the waiting time for the next call follows exponential distribution with expected time $3$ minutes. What is the probability that the next call come in $3$ minutes provided that in the past $2$ minutes there have been no calls?

From the given datils we know $\lambda=\dfrac{1}{3}$ so the distribution

$P(X=x)=\frac{1}{3}e^-\frac{x}{3}$

We have to find $P(0<x<3/x>2)=\dfrac{P(2<x<3)}{p(x>2)}=\frac{\int_{2}^3 \frac{1}{3}e^-\frac{x}{3}\,dx}{\int_{2}^\infty \frac{1}{3}e^-\frac{x}{3}\,dx}$

but I am not sure this answer I think we have to find $P(2<x<3)$ Can someone help me to understand?

Answer 1

The numerator of the fraction is $P(0<x<3 \cap x>2)$. Thus you have to find the intersection of both. A number line helps to find it.

So the required probability is indeed $P(2<x<3)$. Thus the bounds are $2$ and $3$.

Answer 2

From the way you phrased the question, I'm assuming that if the current time is 2pm, you are asking for the probability that a call will come before 2:03 given that no call has been received since 1:58.

The key here is that the exponential distribution is memoryless, i.e. $$\mathbb P(X>a+b\mid X\geq b)=\mathbb P(X>a).$$ So in this case, $$\mathbb P(X>3+2\mid X\geq2)=\mathbb P(X>3)=e^{-\frac{1}{3}\times3}=e^{-1}.$$ So the probability that a customer will arrive in the next three minutes is $1-e^{-1}$. The memoryless property means that the fact we haven't received a call in the past two minutes doesn't affect the probability.

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If instead the interpretation is that the current time is 2pm, no call has been received since 1:58, what is the probability a call will come before 2:01, then the memoryless property implies that the answer is $\mathbb P(X\leq1)=1-e^{-1/3}$.