Calculando Probabilidades: Asistencia A Una Conferencia
Hey guys! Let's dive into a fun probability problem. Imagine a super cool research institute sending out invites for a conference. We're told that, on average, about 60% of the invited guests actually show up. Now, let's say they randomly select 15 guests. Our mission, should we choose to accept it, is to figure out the probabilities of some specific scenarios. Specifically, we're going to calculate the probability that: (i) All 15 guests attend. This is where it gets interesting! This involves understanding binomial probability and how to apply it. We'll break down each part step-by-step so it's super easy to follow. Don't worry if you're not a math whiz; we'll make sure it's clear as day. Ready? Let's roll!
(i) Todos los invitados asistan a la conferencia
Alright, so our first challenge is figuring out the odds that every single one of the 15 guests makes it to the conference. This involves understanding the principles of binomial probability. In this case, each guest either attends (success) or doesn't attend (failure). Since the events are independent, meaning one guest's decision doesn't affect another's, we can use the binomial probability formula. The binomial probability formula is:
P(X = k) = (nCk) * p^k * (1-p)^(n-k)
Where:
P(X = k)is the probability of exactly k successes.nis the number of trials (in our case, the number of guests, which is 15).kis the number of successes (the number of guests attending, which is also 15 in this scenario).pis the probability of success on a single trial (the probability that a guest attends, which is 0.60).nCkis the binomial coefficient, also written as (n choose k), and it's calculated as n! / (k! * (n-k)!).
So, to calculate the probability that all 15 guests attend (k = 15), we plug the numbers into the formula: P(X = 15) = (15C15) * (0.60)^15 * (1-0.60)^(15-15). First, let's calculate (15C15). This is 15! / (15! * 0!), which equals 1. Next, calculate (0.60)^15, which is approximately 0.00047. Finally, calculate (1-0.60)^(15-15) = (0.40)^0, which equals 1. So, P(X = 15) = 1 * 0.00047 * 1 = 0.00047. Therefore, the probability that all 15 guests attend the conference is approximately 0.00047, or about 0.047%. This tells us that it's quite unlikely that all the guests will show up.
Now, let's break this down further. The binomial coefficient (nCk) represents the number of ways to choose k successes from n trials. When k equals n, there's only one way: every single trial is a success. This is why the binomial coefficient is 1. The term p^k represents the probability of k successes, and (1-p)^(n-k) represents the probability of (n-k) failures. In our case, with every guest attending, we have 15 successes, so we multiply the probability of a guest attending (0.60) by itself 15 times, which is approximately 0.00047. The probability of a guest not attending is 0.40, but since all guests attend, this term becomes (0.40)^0, which is 1, as there are no failures. This helps us to understand how the formula works step by step.
It's important to remember that this calculation assumes that the attendance of each guest is independent. Factors like a widespread illness or a major event happening on the same day could change this, but based on the information provided, this is our most accurate estimate. The low probability emphasizes the variability inherent in these kinds of events; it's rare that every invited person will be able to make it!
(ii) Que asistan exactamente 10 invitados
Okay, guys, let's switch gears and figure out the probability that exactly 10 out of the 15 guests show up at the conference. We're still dealing with the binomial probability, so the same formula applies: P(X = k) = (nCk) * p^k * (1-p)^(n-k). But this time, k (the number of successes – guests attending) is 10. Let's crunch the numbers. First, we need to calculate the binomial coefficient (15C10), which is 15! / (10! * 5!). This works out to be 3003. Next, we calculate the probability of success raised to the power of k: (0.60)^10, which is approximately 0.00605. Then, we calculate the probability of failure raised to the power of (n-k): (0.40)^5, which is approximately 0.01024. Finally, multiply all these numbers together: P(X = 10) = 3003 * 0.00605 * 0.01024 = 0.186. Therefore, the probability that exactly 10 guests attend is approximately 0.186, or 18.6%. This is much higher than the probability of all 15 attending.
So, what does this tell us? Well, the most likely outcome is not that every guest attends, nor that nobody attends. It's somewhere in the middle. The probability of exactly 10 guests attending is significant, highlighting that while we can predict the expected number of attendees, there's a range of possibilities. To understand this, let's quickly review the components. The binomial coefficient (15C10) tells us the number of different ways 10 guests could attend the conference out of 15. The term (0.60)^10 gives us the probability of those 10 guests attending, and (0.40)^5 gives the probability of the other 5 not attending. By multiplying these, we consider all possible combinations where exactly 10 out of 15 people are present. This gives us a more realistic picture of the conference attendance. This shows the value of the binomial distribution, which allows us to find the specific probabilities in these cases, and the calculations are pretty much straightforward.
Moreover, the higher probability for 10 guests attending, compared to all 15, makes sense. It's more probable that some guests won't be able to make it than that everyone will. Factors like scheduling conflicts, travel issues, or personal commitments all contribute to variations in attendance. This variance is precisely what the binomial distribution helps us understand. Also, the binomial distribution is very useful in various real-world scenarios, such as predicting the success rate of a new drug in clinical trials, or in quality control, where we can calculate the probability of defective products in a batch. Knowing these probabilities helps us to prepare accordingly.
(iii) Que asistan a lo sumo 12 invitados
Alright, let's calculate the probability that at most 12 guests will be at the conference. This means we need to find the probability that 0 guests attend, 1 guest attends, 2 guests attend, and so on, all the way up to 12 guests attending. We could do this using the binomial probability formula, but we would have to calculate P(X = 0), P(X = 1), P(X = 2), ..., P(X = 12), and then add all those probabilities together. That's quite a bit of work. Instead, we can use a clever trick. The sum of all probabilities in a probability distribution always equals 1. So, if we know the probabilities of the events we don't want (13, 14, and 15 guests attending), we can subtract those from 1. Therefore, P(at most 12) = 1 - [P(13) + P(14) + P(15)].
Let's calculate P(13), P(14), and P(15) first. We already know P(15) is approximately 0.00047 (from part (i)). For P(13): (15C13) * (0.60)^13 * (0.40)^2 = 105 * 0.000016 * 0.16 = 0.00027. For P(14): (15C14) * (0.60)^14 * (0.40)^1 = 15 * 0.000009 * 0.40 = 0.000054. So, P(at most 12) = 1 - (0.00027 + 0.000054 + 0.00047) = 1 - 0.000794 = 0.999. The probability that at most 12 guests attend is approximately 0.999, or 99.9%. This is a very high probability, which means it is extremely likely that no more than 12 guests will attend.
So, this high probability tells us that it's highly likely that the attendance will be at or below 12 guests. The decision to approach the problem in this way made our lives much easier, rather than calculating each individual probability from 0 to 12 and then adding them all up. Understanding this approach helps you to streamline calculations and apply your knowledge of probability. When dealing with