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. What are the probabilities of these events? a. All five operations are successful. b. Exactly four are successful. c. Less than two are successful.

Short answer

Answer: (a) The probability of all five operations being successful is approximately 32.77%. (b) The probability of exactly four operations being successful is approximately 40.96%. (c) The probability of less than two operations being successful is approximately 0.67%.

Step-by-step solution

a. Probability of all five operations being successful

We want to find the probability that all five surgeries are successful, which means \(k=5\). Using the binomial probability formula: \(P(X=5) = \binom{5}{5} (0.8)^5 (1-0.8)^{5-5} = \binom{5}{5} (0.8)^5 (0.2)^0\) Now, calculate the combination term and simplify: \(\binom{5}{5} = \frac{5!}{5!(5-5)!} =\frac{5!}{5!0!} = 1\) Therefore, \(P(X=5) = 1 \times (0.8)^5 \times (0.2)^0 = (0.8)^5 = 0.32768\) The probability that all five operations are successful is approximately 32.77%.

b. Probability of exactly four surgeries being successful

In this case, we want to find the probability that exactly four of the surgeries are successful, which means \(k=4\). Using the binomial probability formula: \(P(X=4) = \binom{5}{4} (0.8)^4 (1-0.8)^{5-4} = \binom{5}{4} (0.8)^4 (0.2)^1\) Now, calculate the combination term and simplify: \(\binom{5}{4} = \frac{5!}{4!(5-4)!} =\frac{5!}{4!1!} = 5\) Therefore, \(P(X=4) = 5 \times (0.8)^4 \times (0.2)^1 = 5 \times 0.4096 \times 0.2 = 0.4096\) The probability of exactly four surgeries being successful is approximately 40.96%.

c. Probability of less than two surgeries being successful

In this case, we want to find the probability that less than two surgeries (i.e., 0 or 1) are successful. We can use the binomial probability formula for both \(k=0\) and \(k=1\) and then add the probabilities together to get the total probability: \(P(X<2) = P(X=0) + P(X=1)\) First, let's find \(P(X=0)\): \(P(X=0) = \binom{5}{0} (0.8)^0 (1-0.8)^{5-0} = \binom{5}{0} (0.8)^0 (0.2)^5\) Now, calculate the combination term and simplify: \(\binom{5}{0} = \frac{5!}{0!(5-0)!} =\frac{5!}{0!5!} = 1\) Therefore, \(P(X=0) = 1 \times (0.8)^0 \times (0.2)^5 = 0.00032\) Next, let's find \(P(X=1)\): \(P(X=1) = \binom{5}{1} (0.8)^1 (1-0.8)^{5-1} = \binom{5}{1} (0.8)^1 (0.2)^4\) Now, calculate the combination term and simplify: \(\binom{5}{1} = \frac{5!}{1!(5-1)!} =\frac{5!}{1!4!} = 5\) Therefore, \(P(X=1) = 5 \times (0.8)^1 \times (0.2)^4 = 5 \times 0.8 \times 0.0016 = 0.0064\) Now, sum the probabilities for \(k=0\) and \(k=1\): \(P(X<2) = P(X=0) + P(X=1) = 0.00032 + 0.0064 = 0.00672\) The probability of less than two surgeries being successful is approximately 0.67%.

Key concepts

Probability Theory

Understanding probability theory is essential when dealing with questions like the one about the surgical procedure's success rates. Probability theory is the branch of mathematics that deals with the likelihood of events occurring. In probability, an event is the outcome or set of outcomes that we are interested in. In this exercise, the events we are interested in are the success of surgical operations.

The basic principle in probability is to quantify uncertainty. Probabilities are always a number between 0 and 1, where 0 means the event will not happen and 1 means the event is certain. For example, if there's an 80% chance a surgery is successful, it can also be expressed as a probability of 0.8.

Probability can be used in various real-life applications, from the likelihood of rain to understanding risk in medical procedures. It helps us predict outcomes and make informed decisions based on statistical data.

Independent Events

Independent events are crucial when considering the probability of multiple events like consecutive surgeries. Two events are independent if the outcome of one event does not affect the outcome of another.

In our example, if each surgery's success or failure does not affect any other surgery, we consider these surgeries as independent events. This assumption allows us to use the binomial probability formula to calculate the likelihood of multiple surgeries being successful.

For independent events, the probability of both events occurring is the product of their individual probabilities. Thus, if each surgery has a probability of success of 0.8, as it is independent from the others, we compute the combined probability for five successful surgeries as \((0.8)^5\).

Understanding independence is vital because it affects the computations for probabilities significantly, ensuring you calculate accurately under the given conditions.

Combinatorics

Combinatorics forms a backbone in solving complex probability problems, especially when we are interested in the number of ways in which events can occur. It involves counting, arrangement, and combination of objects.

When calculating probabilities, such as determining exactly how many surgeries are successful, we utilize combinations. Combinations allow us to count the number of ways to choose formations of successful surgeries.

The binomial coefficient, represented as \(\binom{n}{k}\), is a key formula in combinatorics used here. It calculates the number of ways to pick \(k\) successes in \(n\) trials. For instance, if we want to calculate the probability of exactly four out of five surgeries being successful, we use \(\binom{5}{4}\). This gives us the number of different ways we can choose four successful surgeries out of five.

By understanding combinatorics, you gain insights into how different outcomes can be arranged or combined, which is fundamental in calculating probabilities in a structured way.