# Thread: [SOLVED] Uniform Distribution over two days

1. ## [SOLVED] Uniform Distribution over two days

If T is the time in minutes that bob waits at a bus stop
T~U(0,5)

Excuse the poor paint diagram :P

He gets the bus once on monday and once on tuesday.
What is the probability he will wait for less than 4 minutes in total for both days? (both events are independent).

Ive been trying to do it for like 3 hours now, but im none the wiser, my lecteur notes dont have any examples on it.

2. ## Independent events

Remember that for outcomes A and B of two independent events, $\displaystyle p(A \cap B) = p(A) \cdot P(B)$. So, what is the probability $\displaystyle p = p(t<4)$ that on a singe particular day, Bob waits less than 4 minutes? You then can use these to find the probability of it being less than 4 minutes on both days.

--Kevin C.

3. Originally Posted by corned_beef
If T is the time in minutes that bob waits at a bus stop
T~U(0,5)

Excuse the poor paint diagram :P

He gets the bus once on monday and once on tuesday.
What is the probability he will wait for less than 4 minutes in total for both days? (both events are independent).

Ive been trying to do it for like 3 hours now, but im none the wiser, my lecteur notes dont have any examples on it.
Waiting time is $\displaystyle t_t=t_1+t_2$ where the $\displaystyle t$ 's are the waiting times on the two days.

Now we have two possible approaches, we can use the fact that we know
the distribution of the sum (its a triangular distribution extending from $\displaystyle 0$
to $\displaystyle 10$ minutes with peak at $\displaystyle 5$ minutes).

or we can compute it.

If we wait $\displaystyle t_1$ min on the first day then the probability we require is:

$\displaystyle p(t_t<4|t_1) = \int_0^{4-t_1} p(t_2) dp_2=(4-t_1)$

if $\displaystyle t_1<1$ and $\displaystyle 0$ otherwise.

Then the required prob is:

$\displaystyle p(t_t<4)=\int_0^4 p(t_t<4|t_1) p(t_1) dp_1$

RonL