# Uniform distribution

You arrive at a bus stop at 10 O'clock, knowing that the bus will arrive at some time uniformly distributed between $$10$$ and $$10:30$$. what is the probability that you wait longer than $$10$$ minutes? if at $$10:15$$ the bus has not arrived what is the probability that you will have to wait at least an additional $$10$$ minutes?

Im confused as to if im supposed to take $$P( 5 assuming bus arrives at $$10:15$$, or $$P(15 assuming bus arrives at $$10:30$$. Is this even right? please help!

• It seems like you weren't through typing before you posted. Why don't you go ahead and finish giving us your thoughts on the problem, so we can better help you? Commented Oct 5, 2013 at 6:30
• Unless it is inside MathJax delimiters, $<$ registers as the start of an HTML tag and "eats" the rest of the sentence. Commented Jun 26, 2019 at 6:07

Hint: Let $B$ be the time (in minutes) elapsed since 10 o'clock. Then $B$ is uniformly distribute on the interval $[0,30]$ with density $\dfrac{1}{30}$.
The probability that the waiting time is longer than 10 minutes is $P(B ??)$.
$$P(B>10) = \frac{30-10}{30-0} = \frac23.$$
If the bus hasn't arrived at 10:15 (we're given a condition, so think of the definition of conditional probability), the probability that one will have to wait for at least an additional 10 minutes is $P(B?? | B??)$.
$$P(B>25|B>15) = \frac{P(B>25,B>15)}{P(B>15)} = \frac{P(B>25)}{P(B>15)} = \frac{\frac16}{\frac12} = \frac13$$