# Uniform Distribution with Conditional Probability

#### dsnel23

1. The problem statement, all variables and given/known data
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].

2. Relevant equations

3. 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.

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#### lanedance

Homework Helper
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, teh 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
$$P(B|A) = P(A|B)\frac{P(B)}{P(A)} = (1)\frac{1/3}{2/3} = \frac{1}{2}$$

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