Math Problem Statement

  1. A bus arrives at a station every day at a random time between 1:00 P.M. and 1:30 P.M. A person arrives at this station at 1:00 and waits for the bus. If at 1:15 the bus has not yet arrived, what is the probability that the person will have to wait at least an additional 5 minutes?

Solution

We are given that the bus arrives at a random time between 1:00 P.M. and 1:30 P.M. (uniformly distributed), and a person arrives at 1:00 P.M. If the bus hasn’t arrived by 1:15 P.M., we are asked to find the probability that the person will have to wait at least an additional 5 minutes.

Step-by-step solution:

  1. Time interval:
    The bus can arrive at any time between 1:00 P.M. and 1:30 P.M., meaning the bus arrives uniformly on the interval [0,30][0, 30] minutes (where 0 corresponds to 1:00 P.M. and 30 corresponds to 1:30 P.M.).

  2. Conditional time range:
    At 1:15 P.M., we know that the bus has not arrived yet. This means the bus will arrive sometime between 1:15 P.M. and 1:30 P.M., corresponding to the interval [15,30][15, 30] minutes.

  3. Required event:
    We are asked to find the probability that the person will have to wait at least 5 more minutes from 1:15 P.M., i.e., the bus arrives no earlier than 1:20 P.M. This corresponds to the bus arriving in the interval [20,30][20, 30] minutes.

  4. Uniform distribution: Since the bus arrives uniformly in the interval [15,30][15, 30], the probability that it arrives after 1:20 P.M. (i.e., in the interval [20,30][20, 30]) is given by the ratio of the length of the interval [20,30][20, 30] to the length of the interval [15,30][15, 30].

    • The length of the interval [20,30][20, 30] is 3020=1030 - 20 = 10 minutes.
    • The length of the interval [15,30][15, 30] is 3015=1530 - 15 = 15 minutes.
  5. Final calculation: The probability is:

    [ \text{Probability} = \frac{\text{Length of the interval [20, 30]}}{\text{Length of the interval [15, 30]}} = \frac{10}{15} = \frac{2}{3}. ]

Thus, the probability that the person will have to wait at least an additional 5 minutes after 1:15 P.M. is 23\boxed{\frac{2}{3}}.


Would you like further details on any step or have additional questions?

Here are five related questions to explore:

  1. What is the probability that the bus arrives between 1:10 P.M. and 1:20 P.M.?
  2. How would the probability change if the bus were to arrive between 1:00 P.M. and 1:45 P.M. instead?
  3. If the person arrives at 1:05 P.M., what is the probability that they will wait less than 10 minutes?
  4. What is the expected waiting time for the person if they arrive at 1:00 P.M.?
  5. If the person arrives randomly between 1:00 P.M. and 1:30 P.M., what is the probability that they will have to wait more than 10 minutes?

Tip: When dealing with uniform probability distributions, the probability of an event is often simply the ratio of the favorable interval length to the total possible interval length.

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Math Problem Analysis

Mathematical Concepts

Probability
Uniform Distribution
Conditional Probability

Formulas

P(Event) = Length of favorable interval / Length of total interval

Theorems

Law of Total Probability
Basic Properties of Uniform Distribution

Suitable Grade Level

Grades 9-11