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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 $P( 5<x<10)$ assuming bus arrives at $10:15$, or $P(15<x<20)$ assuming bus arrives at $10:30$. Is this even right? please help!

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    $\begingroup$ 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? $\endgroup$
    – Cameron Buie
    Commented Oct 5, 2013 at 6:30
  • $\begingroup$ Unless it is inside MathJax delimiters, $<$ registers as the start of an HTML tag and "eats" the rest of the sentence. $\endgroup$
    – Graham Kemp
    Commented Jun 26, 2019 at 6:07

1 Answer 1

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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 $P(B?? | B??)$.

$$P(B>25|B>15) = \frac{P(B>25,B>15)}{P(B>15)} = \frac{P(B>25)}{P(B>15)} = \frac{\frac16}{\frac12} = \frac13$$

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