Calculating The Probability Of Reece Arriving At Work By 9 AM

Introduction: Understanding Probability in Everyday Scenarios

In the realm of probability, we often encounter situations where multiple events influence an outcome. Understanding how to calculate these probabilities is crucial in various fields, from business and finance to everyday decision-making. This article will delve into a specific scenario involving Reece's commute to work, where the probability of his bus arriving on time and the subsequent probability of him reaching work by 9 AM are intertwined. By breaking down the problem step-by-step, we'll learn how to calculate the overall probability of Reece arriving at work on time, considering both the scenarios of the bus being on time and the bus being late. This exploration will not only enhance our understanding of probability but also demonstrate its practical application in real-life situations. The principles discussed here can be applied to a wide range of scenarios, making this a valuable exercise in probabilistic thinking.

Problem Statement: Reece's Commute and Probability

Let's consider a scenario where Reece's commute to work involves a bus journey. The probability of Reece's bus arriving on time is given as $\frac{3}{4}$. This forms the foundation of our problem, setting the stage for two distinct possibilities: the bus is on time, or the bus is late. Now, if the bus is indeed on time, the probability of Reece arriving at work by 9 AM is $\frac{2}{3}$. This indicates that even if the bus adheres to its schedule, there's still a chance that unforeseen circumstances might delay Reece's arrival. On the other hand, if the bus is late, the probability of Reece arriving at work by 9 AM drops to $\frac{3}{10}$. This suggests that a late bus significantly impacts Reece's chances of reaching work on time, although it's not impossible. The core question we aim to answer is: What is the overall probability that Reece arrives at work by 9 AM, considering both scenarios – the bus being on time and the bus being late? To solve this, we'll employ the principles of conditional probability and the law of total probability, which will allow us to combine these individual probabilities into a comprehensive answer.

Breaking Down the Probabilities: Defining Events and Outcomes

To effectively tackle this problem, we need to define the events and outcomes clearly. Let's denote the event of Reece's bus arriving on time as 'B' and the event of Reece arriving at work by 9 AM as 'A'. This notation helps us to formalize the given probabilities. We are given that the probability of the bus being on time, denoted as P(B), is $\frac{3}{4}$. Consequently, the probability of the bus being late, denoted as P(B'), is the complement of P(B), which is 1 - $\frac{3}{4}$ = $\frac{1}{4}$. Next, we are given conditional probabilities. The probability of Reece arriving at work by 9 AM given that the bus is on time is P(A|B) = $\frac{2}{3}$. Similarly, the probability of Reece arriving at work by 9 AM given that the bus is late is P(A|B') = $\frac{3}{10}$. These conditional probabilities are crucial because they link the bus's punctuality directly to Reece's arrival time. With these events and probabilities clearly defined, we can now proceed to calculate the overall probability of Reece arriving at work by 9 AM. This involves considering both scenarios – the bus being on time and the bus being late – and their respective probabilities of leading to Reece's timely arrival.

Applying the Law of Total Probability: Calculating the Overall Probability

To determine the overall probability of Reece arriving at work by 9 AM, we employ the law of total probability. This fundamental principle in probability theory allows us to calculate the probability of an event (in this case, Reece arriving at work by 9 AM) by considering all possible scenarios that can lead to that event. In our scenario, there are two such possibilities: the bus is on time, and the bus is late. The law of total probability states that: P(A) = P(A|B) * P(B) + P(A|B') * P(B'), where P(A) is the probability of event A occurring, P(A|B) is the probability of event A occurring given that event B has occurred, P(B) is the probability of event B occurring, P(A|B') is the probability of event A occurring given that event B' (the complement of B) has occurred, and P(B') is the probability of event B' occurring. Now, let's plug in the values we defined earlier: P(A) = ($\frac{2}{3}$ * $\frac{3}{4}$) + ($\frac{3}{10}$ * $\frac{1}{4}$). This equation represents the sum of the probabilities of Reece arriving on time via each scenario. The first term, ($\frac{2}{3}$ * $\frac{3}{4}$), represents the probability of the bus being on time and Reece arriving by 9 AM. The second term, ($\frac{3}{10}$ * $\frac{1}{4}$), represents the probability of the bus being late and Reece still managing to arrive by 9 AM. By calculating this sum, we will obtain the overall probability of Reece arriving at work by 9 AM.

Calculation and Solution: Determining the Final Probability Value

Having established the equation using the law of total probability, we now proceed with the calculation to determine the final probability value. Recall the equation: P(A) = ($\frac{2}{3}$ * $\frac{3}{4}$) + ($\frac{3}{10}$ * $\frac{1}{4}$). First, let's compute each term separately. The first term, ($\frac{2}{3}$ * $\frac{3}{4}$), simplifies to $\frac{6}{12}$, which further reduces to $\frac{1}{2}$. This represents the probability of Reece arriving on time when the bus is on time. The second term, ($\frac{3}{10}$ * $\frac{1}{4}$), simplifies to $\frac{3}{40}$. This represents the probability of Reece arriving on time even when the bus is late. Now, we add these two probabilities together: P(A) = $\frac{1}{2}$ + $\frac{3}{40}$. To add these fractions, we need a common denominator, which is 40. Converting $\frac{1}{2}$ to a fraction with a denominator of 40, we get $\frac{20}{40}$. Therefore, P(A) = $\frac{20}{40}$ + $\frac{3}{40}$ = $\frac{23}{40}$. This final value, $\frac{23}{40}$, represents the overall probability that Reece arrives at work by 9 AM, considering both scenarios of the bus being on time and the bus being late. This result provides a comprehensive understanding of Reece's chances of punctuality, taking into account the various factors influencing his commute.

Conclusion: Interpreting the Results and Implications

In conclusion, by applying the principles of conditional probability and the law of total probability, we have successfully calculated the overall probability of Reece arriving at work by 9 AM. The result, $\frac{23}{40}$, signifies that there is a 57.5% chance (since $\frac{23}{40}$ is equal to 0.575) that Reece will reach work on time. This calculation takes into account the probability of the bus being on time ($\frac{3}{4}$) and the respective probabilities of Reece arriving by 9 AM if the bus is on time ($\frac{2}{3}$) or late ($\frac{3}{10}$). This example demonstrates the power of probability in analyzing real-world scenarios involving multiple events and uncertainties. Understanding these concepts allows us to make informed decisions and predictions based on the likelihood of different outcomes. The same approach can be applied to various other situations, such as predicting market trends, assessing risks in business ventures, or even planning daily activities. The key takeaway is that by breaking down complex problems into smaller, manageable probabilities and applying the appropriate formulas, we can gain valuable insights and make more accurate assessments.

This problem also highlights the importance of considering all possible scenarios when calculating probabilities. Simply focusing on the probability of the bus being on time and Reece arriving by 9 AM if the bus is on time would have provided an incomplete picture. By also considering the scenario where the bus is late, we obtained a more realistic and comprehensive understanding of Reece's chances of arriving at work on time. This underscores the significance of a holistic approach in probability analysis, where all relevant factors and their interactions are taken into account.

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