Uniform Distribution with Conditional Probability

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SUMMARY

The discussion centers on calculating probabilities related to a uniformly distributed waiting time for a bus, specifically using conditional probability. The user correctly calculated the probability of waiting at least 20 minutes as P(X ≥ 20) = 2/3. However, they incorrectly approached Part B, which involved conditional probability after learning that another person had already waited for 20 minutes. The correct calculation for Part B, considering the new information, results in P(X ≥ 20 | waiting 20 minutes) = 1/2, confirming the use of Bayes' theorem in this context.

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Homework Statement


So I just took a probability test and I'm having a hard time with the fact that my answer is wrong. I've done some research online and I believe I am correct, I was hoping to get some input. I'm new to using LaTeX so sorry if it's sloppy. Thanks!

Problem: Suppose that the amount of time in minutes that I wait for a bus is uniformly distributed on the interval [0,60].

Homework Equations


The Attempt at a Solution


(I got part A correct)
Part A: What is the probability I have to wait for at least 20 minutes?

Solution:
P(X \geq 20 ) = 1 - P(X \leq 20) = 1 - \int_0^{20} \frac{1}{60} dx = 1 - \frac{20}{60} = 1 - \frac{1}{3} = \frac{2}{3}

(I got part B wrong)
Part B: Suppose that when I arrive at the bus stop, I meet someone else who has already been waiting for the same bus for 20 minutes. If I take that information into account, what is the probability that I will have to wait for at least 20 minutes?

Solution:
P(X \geq 20 ) = 1 - P(X \leq 20) = 1 - \int_0^{20} \frac{1}{40} dx = 1 - \frac{20}{40} = 1 - \frac{1}{2} = \frac{1}{2}

Since someone else has already been waiting for the same bus for 20 minutes and has not been picked up yet, I would think we could change the sample space? So I used an interval of length 40, since the original was 60.

P.S. I used this site:

http://cnx.org/content/m16819/latest/

Example 2, Problem 3.
 
Last edited by a moderator:
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This might be a good one to use conditional probabilities related by Bayes rule:
P(B|A) = P(A|B)\frac{P(B)}{P(A)}

Consider the first person waiting at the bus stop, the problem translates into:
what is the prob of them waiting for 40min given they have already waited 20min?

Let A be the event X>20, and B X>40
P(A) = \frac{2}{3} and P(B) = \frac{1}{3}

so we want to find P(B|A)
P(B|A) = P(A|B)\frac{P(A)}{P(B)}

clearly P(A|B) = 1, if X>40, then it follows X>20, so plugging in the numbers
P(B|A) = P(A|B)\frac{P(B)}{P(A)} = (1)\frac{1/3}{2/3} = \frac{1}{2}

which agrees with your answer & seems to make sense...
 
Last edited:

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