A new surgical procedure is successful with probability 0.6. Assume the procedure is performed five times and that the results are independent of one another. What is the probability that two or more procedures are successful?.
A new surgical procedure is successful with probability 0.6. Assume the procedure is performed five times and that the results are independent of one another. What is the probability that two or more procedures are successful?.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![### Probability of Surgical Procedure Success
A new surgical procedure is successful with a probability of 0.6. Assume the procedure is performed five times and that the results are independent of one another. What is the probability that two or more procedures are successful?
To solve this problem, we can use the binomial probability formula. The binomial distribution gives the probability of getting exactly \( k \) successes in \( n \) independent Bernoulli trials, each with success probability \( p \).
**Formula**:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Where:
- \( n = 5 \) (number of trials)
- \( p = 0.6 \) (probability of success on each trial)
We need to find the probability of having two or more successes (i.e., \( X \geq 2 \)).
**Steps**:
1. Calculate the probabilities for 0 and 1 success:
- \( P(X = 0) \): No procedures are successful.
- \( P(X = 1) \): Exactly one procedure is successful.
2. Use the complement rule:
\[
P(X \geq 2) = 1 - [P(X = 0) + P(X = 1)]
\]
**Explanation of Binomial Coefficient**:
The binomial coefficient \( \binom{n}{k} \), also known as "n choose k," represents the number of ways to choose \( k \) successes out of \( n \) trials.
Understanding these concepts and calculations will help students grasp the fundamentals of probability related to independent events and binomial distributions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff21aecd2-9b55-44fe-9108-01ab539cadf4%2F41847130-af55-4575-a45c-c9c7355e8333%2Fwu05sbd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability of Surgical Procedure Success
A new surgical procedure is successful with a probability of 0.6. Assume the procedure is performed five times and that the results are independent of one another. What is the probability that two or more procedures are successful?
To solve this problem, we can use the binomial probability formula. The binomial distribution gives the probability of getting exactly \( k \) successes in \( n \) independent Bernoulli trials, each with success probability \( p \).
**Formula**:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Where:
- \( n = 5 \) (number of trials)
- \( p = 0.6 \) (probability of success on each trial)
We need to find the probability of having two or more successes (i.e., \( X \geq 2 \)).
**Steps**:
1. Calculate the probabilities for 0 and 1 success:
- \( P(X = 0) \): No procedures are successful.
- \( P(X = 1) \): Exactly one procedure is successful.
2. Use the complement rule:
\[
P(X \geq 2) = 1 - [P(X = 0) + P(X = 1)]
\]
**Explanation of Binomial Coefficient**:
The binomial coefficient \( \binom{n}{k} \), also known as "n choose k," represents the number of ways to choose \( k \) successes out of \( n \) trials.
Understanding these concepts and calculations will help students grasp the fundamentals of probability related to independent events and binomial distributions.
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