A medicine is known to be 75% effective to cure a patient. If the medicine is given to 5 patients, what is the probability that at least one patient is cured by this medicine?
Understanding Probability of Medicine Effectiveness
This problem asks for the probability that at least one patient is cured by a medicine with a known effectiveness rate. We are given the probability of success for a single patient and the total number of patients. This scenario fits the characteristics of a binomial probability distribution.
Identifying the Problem Parameters
In this case, we can define:
- Success (S): The medicine cures a patient.
- Failure (F): The medicine does not cure a patient.
The given information is:
- Probability of success, \(p = 75\% = \frac{75}{100} = 0.75\).
- Probability of failure, \(q = 1 - p = 1 - 0.75 = 0.25\).
- Number of trials (patients), \(n = 5\).
We want to find the probability that at least one patient is cured. This means the number of cured patients, let's call it \(X\), is 1 or more (\(X \ge 1\)).
Using the Complement Rule
Calculating the probability of "at least one" success directly involves summing the probabilities for \(X=1, X=2, X=3, X=4,\) and \(X=5\). A simpler approach is to use the complement rule of probability:
\[ P(\text{at least one success}) = 1 - P(\text{no successes}) \] In this problem, "no successes" means that none of the 5 patients are cured. This corresponds to \(X=0\).
Calculating Probability of No Patients Cured
The probability of exactly \(k\) successes in \(n\) trials for a binomial distribution is given by the formula:
\[ P(X=k) = \binom{n}{k} p^k q^{n-k} \] We need to find \(P(X=0)\) with \(n=5\), \(k=0\), \(p=0.75\), and \(q=0.25\):
\[ P(X=0) = \binom{5}{0} (0.75)^0 (0.25)^{5-0} \] Let's break down the components:
- \(\binom{5}{0}\) is the number of ways to choose 0 successes from 5 trials. \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), so \(\binom{5}{0} = \frac{5!}{0!(5-0)!} = \frac{5!}{0!5!} = 1\) (since \(0! = 1\)).
- \((0.75)^0 = 1\) (any non-zero number raised to the power of 0 is 1).
- \((0.25)^5\). We can write \(0.25\) as \(\frac{1}{4}\). So, \((0.25)^5 = \left(\frac{1}{4}\right)^5 = \frac{1^5}{4^5}\).
Calculating \(4^5\):
- \(4^1 = 4\)
- \(4^2 = 4 \times 4 = 16\)
- \(4^3 = 16 \times 4 = 64\)
- \(4^4 = 64 \times 4 = 256\)
- \(4^5 = 256 \times 4 = 1024\)
So, \((0.25)^5 = \frac{1}{1024}\).
Now, substitute these values back into the formula for \(P(X=0)\):
\[ P(X=0) = 1 \times 1 \times \frac{1}{1024} = \frac{1}{1024} \] This is the probability that none of the 5 patients are cured by the medicine.
Calculating Probability of At Least One Patient Cured
Using the complement rule:
\[ P(\text{at least one patient cured}) = 1 - P(\text{none cured}) \] \[ P(X \ge 1) = 1 - P(X=0) \] \[ P(X \ge 1) = 1 - \frac{1}{1024} \] To subtract, find a common denominator:
\[ 1 = \frac{1024}{1024} \] \[ P(X \ge 1) = \frac{1024}{1024} - \frac{1}{1024} = \frac{1024 - 1}{1024} = \frac{1023}{1024} \] The probability that at least one of the 5 patients is cured by the medicine is \(\frac{1023}{1024}\).
| Event | Probability |
|---|---|
| Medicine cures a patient (Success, \(p\)) | 0.75 or \(\frac{3}{4}\) |
| Medicine does not cure a patient (Failure, \(q\)) | 0.25 or \(\frac{1}{4}\) |
| Number of patients (\(n\)) | 5 |
| Probability of none cured (\(P(X=0)\)) | \(\left(\frac{1}{4}\right)^5 = \frac{1}{1024}\) |
| Probability of at least one cured (\(P(X \ge 1)\)) | \(1 - P(X=0) = 1 - \frac{1}{1024} = \frac{1023}{1024}\) |
Conclusion on Medicine Effectiveness Probability
The probability that at least one patient is cured when the medicine is given to 5 patients, with a 75% effectiveness rate per patient, is \(\frac{1023}{1024}\).
Revision Table: Probability Concepts
| Concept | Description | Formula Example |
|---|---|---|
| Probability of Success (\(p\)) | The likelihood of a desired outcome in a single trial. | Given as 75% or 0.75 |
| Probability of Failure (\(q\)) | The likelihood of the outcome not being a success. \(q = 1 - p\). | \(1 - 0.75 = 0.25\) |
| Binomial Probability | Probability of exactly \(k\) successes in \(n\) independent trials. | \(P(X=k) = \binom{n}{k} p^k q^{n-k}\) |
| Complement Rule | The probability of an event not happening is 1 minus the probability of the event happening. | \(P(A') = 1 - P(A)\) |
| At Least One Probability | Probability of 1 or more successes. Often calculated as \(1 - P(\text{zero successes})\). | \(P(X \ge 1) = 1 - P(X=0)\) |
Additional Information on Binomial Distribution and Probability
The binomial distribution is used for experiments that have the following characteristics:
- A fixed number of trials (\(n\)). In this problem, \(n=5\).
- Each trial is independent. The outcome for one patient does not affect others.
- There are only two possible outcomes for each trial: success (cured) or failure (not cured).
- The probability of success (\(p\)) is constant for each trial. In this problem, \(p=0.75\) for every patient.
The complement rule is a very useful tool in probability. It simplifies calculations for events like "at least one" by allowing us to calculate the probability of the opposite event (none) and subtract it from 1. This is particularly helpful when the number of trials is large, as it avoids summing many individual probabilities.
In this specific problem, \(\binom{5}{0}\) represents the number of ways to choose which of the 5 patients are cured if none are cured. There is only 1 way for none to be cured (i.e., all fail), which aligns with \(\binom{5}{0}=1\).
Similar Questions
Let X and Y be two random variables such that X + Y = 100. If X follows Binomial distribution with parameters n = 100 and p = \(\frac{4}{5}\), what is the variance of Y?
In a Binomial distribution B(n, p), n = 6 and 9P(X = 4) = P(X = 2). What is p equal to ?
What is the probability that no one out of 6 workers suffers from a disease?
What is the probability that exactly 3 out of 6 workers suffer from a disease?
What is the probability that at least one out of 6 workers suffer from a disease ?
8 coins are tossed simultaneously. The probability of getting at least 6 heads is
A certain type of missile hits the target with probability p = 0.3. What is the least number of missiles should be fired so that there is at least on 80% probability that the target is hit?
Let the random variables X follow B (6, p) If \(16{\rm{\;P\;}}\left( {{\rm{X}} = 4} \right) = {\rm{P\;}}\left( {{\rm{X}} = 2} \right)\) , then what is the value of p
The mean and the variance in a binomial distribution are found to be 2 and 1 respectively. The probability P(X = 0) is
In an examination, the probability of a candidate solving a question is 1/2. out of given 5 questions in the examination, what is the probability that the candidate was able to solve at least 2 questions?
Important Questions from Binomial Distribution
Indicate the correct answer for the combination from the following regarding the conditions for the applicability of a binominal distribution:
(a) There are n independent trials
(b) Each trial has only two possible outcomes
(c) The probabilities of two outcomes do not remain constant
(d) The trials are independent
Which of the following options is correct?
In which of the following practical situations, Poisson Distribution can be used?
A. Number of customers arriving at the super markets per hour.
B. Number of typographical errors per page in a typed material.
C. Number of accidents taking place per day on a busy road.
D. Dice throwing problems.
E. Number of defective material say, blades, etc. in a packing manufactured by a good concern.
Choose the most appropriate answer from the options given below:
For Binomial distribution, n = 10 and p = 0.6, E(X 2) (second moment about origin) is:
The mean and variance of binomial distribution B (x, n, p) are 4 and \(\dfrac{4}{3}\) respectively. What is the probability of getting 2 successes?
Find out the fallacy if any in the statement:
“The mean and the variance of a binomial distribution is 16.2 and 29.4 respectively.”
