# Waiting time in a merged queue described by a random variable with exponential distribution.

I have a problem with merging 2 exponential distributions.

Let's say that: people need to wait in a queue $$1$$ and then wait in queue $$2$$ or queue $$3$$ (with probability $$\frac{1}{2}$$ for each).

1. Waiting time in queue $$1$$ is a random variable with an exponential distribution, mean of $$\frac{1}{3}$$ hour.
2. Waiting time in queue $$2$$ is a random variable with an exponential distribution, mean of $$\frac{1}{4}$$ hour
3. Waiting time in queue $$3$$ is a random variable with an exponential distribution, mean of $$\frac{1}{5}$$ hour.

All these random variables are independent.

What's the probability that some man will wait for less than $$20$$ minutes?

I would know how to answer when taking the probability for $$1$$ queue ($$1$$ exponential distribution). However, I don't know how to merge them together because a man can wait for a relatively short time in the first one and for a relatively long time in a second. I don't know how to capture it with a description in a form of inequality using both exponential distributions.

• hint... Poisson May 29, 2023 at 4:48

$$P(T<1/3)=\tfrac12 P(T_1+T_2<1/3)+\tfrac12 P(T_1+T_3<1/3).$$
Then use the following result giving the cdf (cumulative distribution function) of the sum of two independent exponential random variables $$X,Y$$ with resp. parameters $$\lambda,\mu$$ :
$$P(X+Y