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

Question

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.

Answer 1

[At first, I had misread your text].

Here is the correct way to answer :

$$P(T<1/3)=\tfrac12 P(T_1+T_2<1/3)+\tfrac12 P(T_1+T_3<1/3).$$

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 $\lambda,\mu$ :

$$P(X+Y<t)=1 - \frac{\mu e^{-\lambda t} -\lambda e^{-\mu t} }{\mu - \lambda}$$

(see here for a proof).