Question

A new surgical procedure is said to be successful \(80 \%\) of the time. Suppose the operation is performed five times and the results are assumed to be independent of one another. a. What is the probability that all five operations are successful. b. What is the probability that exactly four are successful. c. What is the probability that less than two are successful. d. If less than two operations were successful, how would you feel about the performance of the surgical team?

Short answer

Answer: The probability that all five operations are successful is 32.77%, the probability that exactly four are successful is 40.96%, and the probability that less than two are successful is 0.67%.

Step-by-step solution

Identify the Binomial Probability Formula

First, let's identify the binomial probability formula, which is the formula that will help us find the probability of different outcomes when we have a fixed number of independent trials with the same probability of success. The binomial probability formula is: P(X=k) = C(n,k) * p^k * (1-p)^(n-k) where: - P(X=k) is the probability of exactly k successes in n trials - C(n,k) is the combination formula which represents the number of ways to choose k successes out of n trials (C(n,k)= n! / (k!(n-k)!)) - n is the number of trials - k is the number of successes - p is the probability of success in a single trial (in this case, 0.8)

Setup for the problem

Assuming each surgery is an independent trial with probability \(p = 0.8\) for each surgery, we have the components we need for our formula. Let's define the values for our problem: - n = 5 (total number of surgeries) - p = 0.8 (probability of successful surgery) - k = will vary depending on the question part

Calculate the probabilities for each part

a) Probability that all five operations are successful: - k = 5 (five successful operations) - P(X=5) = C(5,5) * 0.8^5 * (1-0.8)^(5-5) - P(X=5) = 1 * 0.32768 * 1 - P(X=5) = 0.32768 => 32.77% b) Probability that exactly four are successful: - k = 4 (four successful operations) - P(X=4) = C(5,4) * 0.8^4 * (1-0.8)^(5-4) - P(X=4) = 5 * 0.4096 * 0.2 - P(X=4) = 0.4096 => 40.96% c) Probability that less than two are successful (either 0 or 1 successful surgery): - k = 0 (zero successful operations) - P(X=0) = C(5,0) * 0.8^0 * (1-0.8)^(5-0) - P(X=0) = 1 * 1 * 0.00032 - P(X=0) = 0.00032 => 0.03% - k = 1 (one successful operation) - P(X=1) = C(5,1) * 0.8^1 * (1-0.8)^(5-1) - P(X=1) = 5 * 0.8 * 0.0016 - P(X=1) = 0.0064 => 0.64% - Probability that less than two are successful = P(X=0) + P(X=1) = 0.00032 + 0.0064 = 0.00672 => 0.67%

Interpret the results

a) The probability that all five operations are successful is 32.77%, meaning there is a 32.77% chance that all five surgeries will be successful. b) The probability that exactly four are successful is 40.96%, meaning there is a 40.96% chance that four out of the five surgeries will be successful. c) The probability that less than two are successful is 0.67%, meaning there is a 0.67% chance that either none or only one surgery out of the five will be successful. d) If less than two operations were successful (probability of 0.67%), it would indicate that the surgical team might have performed well below their expected 80% success rate, and it might be worth investigating the reasons behind such a poor outcome. However, it is important to note that probabilities for each individual surgery were independent, so it could also be a rare unfortunate event.

Key concepts

Probability of Success

Understanding the probability of success is pivotal when dealing with binomial probability. In our context, it’s defined as the likelihood of a single trial resulting in a favorable outcome—in this case, a successful surgical procedure. To put it in perspective, imagine flipping a coin with a biased 80% chance of landing heads. Here, heads would represent a successful surgery, and tails a failure.

With a probability of success of 0.8, we expect, on average, eight successes in ten trials. But, this is an average expectation—not a guarantee, and it’s important for students to not conflate high probability with certainty. In our exercise, this notion helps students form expectations about the surgical procedures' outcomes based on mathematical grounds rather than intuition.

The ease of the formula's application lies in its consistency; whether we're considering one operation or a sequence of them, the success probability remains constant at 0.8 for each independent trial. This consistency is what qualifies a process for analysis via binomial probability.

Independent Trials

Independent trials are the cornerstone of the binomial probability model. When we state that trials are independent, we mean that the outcome of one does not influence the outcome of another. A classic example is rolling a die: no matter what appears in the first roll, the second roll is unaffected.

In our surgical case, it’s crucial to bear in mind that the success or failure of each surgery does not depend on the results of the others to apply the binomial probability formula accurately. This is a common source of confusion and error in probability calculations, so it's vital to guarantee independence before proceeding.

Importance of Independence

Should any two surgeries be linked, perhaps if one patient's outcome directly influenced another’s, the calculations and resulting probabilities would change entirely, and the model would not apply. It's this independence that allows us to use the formula without adjustments for each scenario, be it all operations successful, exactly four successful, or less than two successful.

Combinatorial Mathematics

Combinatorial mathematics allows us to understand and quantify the number of ways in which a certain number of elements can be selected or arranged. In the realm of binomial probability, combinatorial mathematics helps us figure out how many different ways we can achieve a set number of successes in a series of trials.

The combination formula, denoted as C(n, k), is used in our surgery example to calculate how many ways we can get zero, one, four, or five successful operations out of five. The factorial operation (!) plays a critical role here, representing the product of all positive integers up to a given number.

Combinations in Real Life

For example, when considering exactly four successful surgeries, C(5,4) translates to 'in how many ways can we pick four items from a set of five?' which is relevant to real-world situations like lottery drawings or determining genetic probabilities.

This combinatorial aspect is integral in calculating probabilities for each part of our exercise. It is not only a theoretical construct but also a practical tool when analyzing both simple and complex probability scenarios.