# 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? – Cameron Buie Oct 5 '13 at 6:30
• Unless it is inside MathJax delimiters, $<$ registers as the start of an HTML tag and "eats" the rest of the sentence. – Graham Kemp Jun 26 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 $P(B?? | B??)$.
$$P(B>25|B>15) = \frac{P(B>25,B>15)}{P(B>15)} = \frac{P(B>25)}{P(B>15)} = \frac{\frac16}{\frac12} = \frac13$$