Problem
The taxi stand and the bus stop near AI's home are in the same location. Al goes there at a given time and if a taxi is waiting (this happens with probability 2/3) he boards it. Otherwise he waits for a taxi or a bus to come, whichever comes first. The next taxi will arrive in a time that is uniformly distributed between 0 and 10 minutes, while the next bus will arrive in exactly 5 minutes. Find the CDF and the expected value of AI's waiting time.
Answer
Solution. Let A be the event that Al will find a taxi waiting or will be picked up by the bus after 5 minutes. Note that the probability of boarding the next bus, given that Al has to wait, is
$$
P(\text{taxi will take more than 5 minutes to arrive}) = \frac{1}{2}.
$$
…
My problem
Why that probability is 1/2? if the $P(\text{first 5min no taxi})=\frac{1}{3}$ and $P(\text{then catch the taxi})=\frac{2}{3}$, then $P(\text{taxi will take more than 5 minutes to arrive}) = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$. But why it's $\frac{1}{2}$
Best Answer
The probability that the taxi takes more than $5$ mins to arrive is the same as $P(U\geq 5)$ for an uniform$[0,10]$ variate which is same as $\frac{5}{10}=\frac{1}{2}$
The way you do it is the following,
Let $X$ be the waiting time for Al.
Then $P(X=0)=\frac{2}{3}$ as given.
Now see that $P(X=x+dx)$ where $0<x<5$ is given by $\frac{1}{3}\cdot\frac{1}{10}$, i.e. there was no taxi in the beginning and Al waits $x$ minutes which is just $\frac{1}{3}$ times the pdf of the uniform$[0,10]$ variate. Alternatively, see that for $0<x<5$, you have: \begin{align}P(X\leq x)&=P(X=0)+P(X\leq x,X\neq 0)\\ &=\frac{2}{3}+P(U\leq x,X\neq 0)\\ &=\frac{2}{3}+P(U\leq x)P(X\neq 0)\,, (\text{by independence})\\ &=\frac{2}{3}+(\int_{0}^{x}\frac{1}{10}\,dx)\cdot\frac{1}{3}\end{align}
And $P(X=5)$ is given by the probability that there was no taxi in the first $5$ minutes and there was no taxi in the beginning which is $\frac{1}{3}\cdot P(U\geq 5)=\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}$
Hence, the cdf is given by $\displaystyle F_{X}(x)=\begin{cases}0\,,x<0\\\frac{2}{3}\,,x=0\\ \frac{2}{3}+\frac{x}{30}\,,0<x<5\\ 1\,,x\geq 5\\ \end{cases}$
Now find the expectation $\displaystyle E(X)=0\cdot \frac{2}{3}+5\cdot\frac{1}{6}+\int_{0}^{5}x\cdot\frac{1}{30}\,dx=\frac{5}{6}+\frac{5}{12}=\frac{5}{4}$