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%
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).