Calculating The Probability Of Abiola Being Late To The Office

Hey guys! Let's dive into a fun probability problem today. We're going to figure out the chances of Abiola being late to the office. This is a classic example of a binomial probability scenario, and we'll break it down step by step so it's super easy to understand. We will discuss the concepts involved and provide a clear, step-by-step solution. We'll also touch upon why these kinds of probability problems are important in real-world situations. So, grab your thinking caps, and let's get started!

The Problem

Okay, here's the problem we're tackling: Abiola has a $2 / 5$ chance of being late to work each day. We're looking at a six-day work week. What's the probability that Abiola will be late for exactly three days? And what's the probability that he won't be late at all during the week? These are the questions we're going to answer, making sure to get those answers accurate to four significant figures. To solve this, we'll use the binomial probability formula, which is perfect for scenarios with two outcomes (late or not late) and a fixed number of trials (six days).

a) Probability of Being Late for Exactly 3 Days

So, let's tackle the first part of the problem: what's the probability that Abiola is late for exactly three days out of the six-day work week? This is where the binomial probability formula comes into play. The binomial probability formula helps us calculate the probability of getting a specific number of successes (in this case, being late) in a fixed number of trials (days), given a constant probability of success on each trial. It's a bit of a mouthful, but it's a powerful tool. Before we jump into the formula, let's define our terms. We'll use n to represent the total number of trials (days), k to represent the number of successful trials we're interested in (days late), p to represent the probability of success on a single trial (probability of being late on a given day), and q to represent the probability of failure (probability of not being late on a given day). In our case, n is 6 (days), k is 3 (days late), p is $2 / 5$ (probability of being late), and q is $1 - p = 3 / 5$ (probability of not being late). Now we have all the pieces we need to plug into the binomial probability formula. The binomial probability formula is: P(X = k) = C(n, k) * p^k * q^(n-k). Where: P(X = k) is the probability of exactly k successes in n trials. C(n, k) is the number of combinations of n items taken k at a time (also known as "n choose k"). p is the probability of success on a single trial. q is the probability of failure on a single trial. Now, let's break down each part of the formula and see how it applies to our problem. C(n, k) represents the number of ways we can choose k successful trials from n total trials. It's calculated using the formula: C(n, k) = n! / (k! * (n-k)!). The exclamation mark (!) denotes the factorial, which means multiplying a number by all the positive integers less than it (e.g., 5! = 5 * 4 * 3 * 2 * 1). So, C(6, 3) = 6! / (3! * 3!) = (6 * 5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (3 * 2 * 1)) = 20. This means there are 20 different ways Abiola can be late for exactly three days out of six. Now, let's move on to the p^k part of the formula. This represents the probability of getting k successes in a row. In our case, it's the probability of Abiola being late for three days in a row. Since the probability of being late on any given day is $2 / 5$, the probability of being late for three days in a row is ($2 / 5$)^3. Next, we have q^(n-k), which represents the probability of getting (n-k) failures. In our case, it's the probability of Abiola not being late for the remaining (6 - 3) = 3 days. Since the probability of not being late on any given day is $3 / 5$, the probability of not being late for three days is ($3 / 5$)^3. Now that we've calculated all the individual components, we can plug them back into the binomial probability formula: P(X = 3) = C(6, 3) * p^3 * q^3 = 20 * ($2 / 5$)^3 * ($3 / 5$)^3 = 20 * (8 / 125) * (27 / 125) = 0.27648. Rounding this to four significant figures, we get 0.2765. So, the probability that Abiola will be late for exactly three days in a six-day work week is approximately 0.2765, or about 27.65%. That's pretty interesting, isn't it? It shows how we can use probability to predict the likelihood of certain events happening.

b) Probability of Not Being Late at All

Now, let's move on to the second part of our problem: what's the probability that Abiola won't be late at all during the six-day work week? This might seem tricky, but we can use the same binomial probability formula we used before! The key here is to recognize that "not being late at all" means being late for zero days. So, in our formula, k will be 0. Remember our other variables? n is still 6 (the number of days in the work week), p is still $2 / 5$ (the probability of being late on any given day), and q is still $3 / 5$ (the probability of not being late on any given day). So, we will recalculate the probability that Abiola is late for zero days out of the six-day work week. Let's use the binomial probability formula again, P(X = k) = C(n, k) * p^k * q^(n-k). We're plugging in our new value for k (0), and keeping the other values the same. This time, we have P(X = 0) = C(6, 0) * ($2 / 5$)^0 * ($3 / 5$)^(6-0). Let's break this down step by step. First, we need to calculate C(6, 0), which is the number of ways to choose 0 days out of 6. Using the formula C(n, k) = n! / (k! * (n-k)!), we get C(6, 0) = 6! / (0! * 6!). Remember that 0! is defined as 1. So, C(6, 0) = 1. This makes sense, right? There's only one way to choose zero days – you simply choose none of them. Next, we have ($2 / 5$)^0. Any number raised to the power of 0 is 1, so ($2 / 5$)^0 = 1. This part is easy! Now, we need to calculate ($3 / 5$)^6, which is the probability of not being late for six consecutive days. This is (0.6)^6 = 0.046656. Finally, we plug all these values back into the formula: P(X = 0) = 1 * 1 * 0.046656 = 0.046656. Rounding this to four significant figures, we get 0.04666. So, the probability that Abiola won't be late at all during the six-day work week is approximately 0.04666, or about 4.67%. This is quite a bit lower than the probability of being late for three days, which makes sense. It's less likely to have a perfect week of punctuality than to have a mix of late and on-time days. It is important to know that these kinds of calculations are not just academic exercises. They have practical applications in various fields, including business, finance, and even healthcare. For instance, businesses might use probability to model customer behavior or predict sales trends. In finance, probability is used to assess risk and make investment decisions. In healthcare, it can be used to analyze the effectiveness of treatments or predict the spread of diseases. So, understanding probability is a valuable skill, no matter what field you're in.

Conclusion

So, guys, we've successfully tackled a probability problem involving Abiola's lateness! We found that the probability of him being late for exactly three days in a six-day work week is approximately 0.2765, and the probability of him not being late at all is approximately 0.04666. We used the binomial probability formula, which is a powerful tool for solving problems with two outcomes and a fixed number of trials. We also discussed why these kinds of probability calculations are important in real-world situations. The key takeaway here is that probability helps us understand and predict the likelihood of events happening. It's a fundamental concept that has wide-ranging applications. And remember, even seemingly complex problems can be broken down into smaller, manageable steps. Just take it one step at a time, and you'll be amazed at what you can accomplish.

I hope this explanation was helpful and easy to follow! If you have any questions, feel free to ask. Keep practicing, and you'll become a probability pro in no time!