buses arrive at a bus stop at 15minute interval,starting from
7am if a passenger arrived at a times that is his uniformly distributed
(a):find the probability that the passenger wait less than 5minute for a bus?
(b):find the probability that the passenger wait more than 10minute for a bus?
thanks Step plz
all wait times have an equal probability
so the probability of being in any one of three five-minute intervals is ⅓
(a) is the 1st interval
(b) is the last interval
To solve this problem, we need to find the probability distribution of the time a passenger has to wait for a bus. Let's break it down step by step:
Step 1: Understand the problem
We are given that buses arrive at a bus stop at 15-minute intervals, starting from 7 am. We are also told that the arrival time of a passenger is uniformly distributed.
Step 2: Define the random variable
Let X be the random variable representing the time a passenger has to wait for a bus.
Step 3: Determine the probability distribution
Since the arrival time of a passenger is uniformly distributed, we know that the probability density function (PDF) is constant between two consecutive bus arrivals and zero elsewhere.
Given that buses arrive at 15-minute intervals, the PDF of X can be defined as:
f(x) = 1/15 if 0 ≤ x < 15
= 0 otherwise
Step 4: Calculate the probability
(a) To find the probability that the passenger waits less than 5 minutes for a bus, we need to calculate P(X < 5).
We can find the probability by integrating the PDF from 0 to 5:
P(X < 5) = ∫[0,5] f(x) dx
Since f(x) is a constant in the interval [0, 5], we can simplify the integral:
P(X < 5) = (1/15) * (5 - 0) = (1/15) * 5 = 1/3
Therefore, the probability that the passenger waits less than 5 minutes for a bus is 1/3.
(b) To find the probability that the passenger waits more than 10 minutes for a bus, we need to calculate P(X > 10).
We can find the probability by integrating the PDF from 10 to 15:
P(X > 10) = ∫[10,15] f(x) dx
Again, since f(x) is a constant in the interval [10, 15], we can simplify the integral:
P(X > 10) = (1/15) * (15 - 10) = (1/15) * 5 = 1/3
However, we need to find the probability of waiting "more" than 10 minutes, so we subtract this value from 1:
P(X > 10) = 1 - P(X < 10) = 1 - (1/3) = 2/3
Therefore, the probability that the passenger waits more than 10 minutes for a bus is 2/3.
In summary:
(a) The probability that the passenger waits less than 5 minutes for a bus is 1/3.
(b) The probability that the passenger waits more than 10 minutes for a bus is 2/3.