Problem 13
Question
You arrive at a bus stop at 10 A.M., knowing that the bus will arrive at some time uniformly distributed between 10 and 10: 30 (a) What is the probability that you will have to wait longer than 10 minutes? (b) If, at \(10: 15,\) the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes?
Step-by-Step Solution
Verified Answer
In summary, the probability that you will have to wait longer than 10 minutes is \(\frac{2}{3}\), and the probability that you will have to wait at least an additional 10 minutes, given that the bus has not yet arrived at 10:15 AM, is \(\frac{1}{3}\).
1Step 1: (a) Determine the probability density function
Firstly, we need to find the probability density function (pdf) of the bus arrival time. Since the bus will arrive at some time uniformly distributed between 10:00 AM and 10:30 AM, we can write the interval as [0, 30] minutes. The pdf of a uniform distribution is given by:
\(f(x)=\frac{1}{b-a}\) for \(a \le x \le b\)
In our case, \(a=0\) and \(b=30\). So, the pdf is:
\(f(x)=\frac{1}{30-0}=\frac{1}{30}\) for \(0 \le x \le 30\)
2Step 2: (a) Calculate the probability of waiting longer than 10 minutes
To find the probability of waiting longer than 10 minutes, we need to calculate the area under the pdf curve from 10 minutes to 30 minutes. We can integrate the pdf function over this interval:
\(P(X>10)=\int_{10}^{30} f(x) dx = \int_{10}^{30} \frac{1}{30} dx\)
Now, we can solve this integral:
\(P(X>10) = \left[ \frac{x}{30} \right]_{10}^{30} = \frac{30}{30} - \frac{10}{30} = \frac{2}{3}\)
So, the probability that you will have to wait longer than 10 minutes is \(\frac{2}{3}\).
3Step 3: (b) Calculate the conditional probability
In this part of the exercise, we are asked to calculate the probability that you will have to wait at least an additional 10 minutes given that the bus has not arrived at 10:15 AM. This is a conditional probability problem. We can represent this probability as \(P(X>25|X>15)\).
Using the properties of conditional probability, we can write:
\(P(X>25|X>15)=\frac{P(X>25 \cap X>15)}{P(X>15)}\)
Since \(X>25\) implies \(X>15\), the probability in the numerator is simply the probability of \(X>25\):
\(P(X>25)=\int_{25}^{30} f(x) dx = \int_{25}^{30} \frac{1}{30} dx\)
Now, we can solve this integral:
\(P(X>25) = \left[ \frac{x}{30} \right]_{25}^{30} = \frac{30}{30} - \frac{25}{30} = \frac{1}{6}\)
Next, we need to find the probability of the bus not arriving until after 10:15 AM:
\(P(X>15)=\int_{15}^{30} f(x) dx = \int_{15}^{30} \frac{1}{30} dx\)
Solve the integral:
\(P(X>15) = \left[ \frac{x}{30} \right]_{15}^{30} = \frac{30}{30} - \frac{15}{30} = \frac{1}{2}\)
Now, we can find the conditional probability:
\(P(X>25|X>15)=\frac{P(X>25)}{P(X>15)}=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1/6}{1/2}=\frac{2}{6}=\frac{1}{3}\)
Thus, the probability that you will have to wait at least an additional 10 minutes, given that the bus has not yet arrived at 10:15 AM, is \(\frac{1}{3}\).
Key Concepts
Probability Density FunctionConditional ProbabilityUniform Distribution
Probability Density Function
Understanding probability requires familiarity with the concept of a probability density function (pdf), which is crucial when we deal with continuous probability distributions. The pdf helps determine the likelihood that a random variable takes on a value within a specific interval.
For a continuous random variable, the probability density function represents how the total probability is distributed across possible values the variable can assume. By definition, it must satisfy two conditions: first, the pdf needs to be non-negative at each point; and second, when integrated over all possible values, it must equal to one, indicating a certainty that the random variable will take on a value within the given range.
A uniform distribution's pdf is particularly straightforward—it has the same value across the entire interval where the variable can take on values, and zero otherwise. The pdf for a uniform distribution is given by \( f(x) = \frac{1}{b-a} \) for \( a \le x \le b \), where the interval \[a, b\] contains all the possible values of the random variable. The simplicity of a uniform distribution's pdf reflects the concept that every outcome within the specified range is equally likely.
For a continuous random variable, the probability density function represents how the total probability is distributed across possible values the variable can assume. By definition, it must satisfy two conditions: first, the pdf needs to be non-negative at each point; and second, when integrated over all possible values, it must equal to one, indicating a certainty that the random variable will take on a value within the given range.
A uniform distribution's pdf is particularly straightforward—it has the same value across the entire interval where the variable can take on values, and zero otherwise. The pdf for a uniform distribution is given by \( f(x) = \frac{1}{b-a} \) for \( a \le x \le b \), where the interval \[a, b\] contains all the possible values of the random variable. The simplicity of a uniform distribution's pdf reflects the concept that every outcome within the specified range is equally likely.
Conditional Probability
Another key concept in understanding probabilities in various scenarios is conditional probability. It's the probability of an event occurring given that another event has already occurred, which can have a significant impact on the outcome.
The conditional probability of an event B given an event A is denoted by \( P(B|A) \), and is calculated by the formula \( P(B|A) = \frac{P(A \cap B)}{P(A)} \) as long as \( P(A) > 0\). Essentially, it tells us how to adjust our expectations for B in light of the fact that A has happened.
Conditional probability can change our understanding of a situation quite dramatically. For example, in the context of waiting for a bus, if you know the bus hasn't arrived in the first 15 minutes, this knowledge updates the likelihood of how much longer you'll need to wait. If we consider the probability that two events occur together, compared to the probability that one occurs on its own, conditional probability provides a framework to quantify this relationship.
The conditional probability of an event B given an event A is denoted by \( P(B|A) \), and is calculated by the formula \( P(B|A) = \frac{P(A \cap B)}{P(A)} \) as long as \( P(A) > 0\). Essentially, it tells us how to adjust our expectations for B in light of the fact that A has happened.
Conditional probability can change our understanding of a situation quite dramatically. For example, in the context of waiting for a bus, if you know the bus hasn't arrived in the first 15 minutes, this knowledge updates the likelihood of how much longer you'll need to wait. If we consider the probability that two events occur together, compared to the probability that one occurs on its own, conditional probability provides a framework to quantify this relationship.
Uniform Distribution
In probability and statistics, a uniform distribution is one of the simplest forms of distributions where all outcomes are equally likely. If a random variable X follows a uniform distribution over an interval \[a, b\], any value within the interval is as likely as any other.
This distribution is characterized by two parameters, the minimum and maximum values \(a\) and \(b\), which define the range of possible outcomes. The uniform distribution is often represented graphically as a rectangle because the probability is perfectly evenly distributed over the interval, hence its alternate name, rectangular distribution.
In real-world applications, uniform distributions are ideal for representing scenarios where a random variable has a finite number of outcomes that are equally probable, such as drawing a random card from a well-shuffled deck or, as our exercise describes, waiting for a bus that is equally likely to arrive at any minute within a half-hour window.
This distribution is characterized by two parameters, the minimum and maximum values \(a\) and \(b\), which define the range of possible outcomes. The uniform distribution is often represented graphically as a rectangle because the probability is perfectly evenly distributed over the interval, hence its alternate name, rectangular distribution.
In real-world applications, uniform distributions are ideal for representing scenarios where a random variable has a finite number of outcomes that are equally probable, such as drawing a random card from a well-shuffled deck or, as our exercise describes, waiting for a bus that is equally likely to arrive at any minute within a half-hour window.
Other exercises in this chapter
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