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?

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).