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

Answer:

a) the probability of waiting more than 10 min is 2/3 ≈ 66,67%

b) the probability of waiting more than 10 min, knowing that you already waited 15 min is 5/15 ≈ 33,33%

Step-by-step explanation:

to calculate, we will use the uniform distribution function:

p(c≤X≤d)= (d-c)/(B-A) , for A≤x≤B

where p(c≤X≤d) is the probability that the variable is between the values c and d. B is the maximum value possible and A is the minimum value possible.

In our case the random variable X= waiting time for the bus, and therefore

B= 30 min (maximum waiting time, it arrives 10:30 a.m)

A= 0 (minimum waiting time, it arrives 10:00 a.m )

a) the probability that the waiting time is longer than 10 minutes:

c=10 min , d=B=30 min --> waiting time X between 10 and 30 minutes

p(10 min≤X≤30 min) = (30 min - 10 min) / (30 min - 0 min) = 20/30=2/3 ≈ 66,67%

a) the probability that 10 minutes or more are needed to wait starting from 10:15 , is the same that saying that the waiting time is greater than 25 min (X≥25 min) knowing that you have waited 15 min (X≥15 min). This is written as P(X≥25 | X≥15 ). To calculate it the theorem of Bayes is used

P(A | B )= P(A ∩ B ) / P(A) . where P(A | B ) is the probability that A happen , knowing that B already happened. And P(A ∩ B ) is the probability that both A and B happen.

In our case:

P(X≥25 | X≥15 )= P(X≥25 ∩ X≥15 ) / P(X≥15 ) = P(X≥25) / P(X≥15) ,

Note: P(X≥25 ∩ X≥15 )= P(X≥25) because if you wait more than 25 minutes, you are already waiting more than 15 minutes

-   P(X≥25) is the probability that waiting time is greater than 25 min

c=25 min , d=B=30 min --> waiting time X between 25 and 30 minutes

p(25 min≤X≤30 min) = (30 min - 25 min) / (30 min - 0 min) = 5/30 ≈ 16,67%

-  P(X≥15) is the probability that waiting time is greater than 15 min --> p(15 min≤X≤30 min) = (30 min - 15 min) / (30 min - 0 min) = 15/30

therefore

P(X≥25 | X≥15 )= P(X≥25) / P(X≥15) = (5/30) / (15/30) =5/15=1/3  ≈ 33,33%

Note:

P(X≥25 | X≥15 )≈ 33,33% ≥ P(X≥25) ≈ 16,67%  since we know that the bus did not arrive the first 15 minutes and therefore is more likely that the actual waiting time could be in the 25 min - 30 min range (10:25-10:30).