The probability that a person recovers from a disease is 0.8. What is the probability that exactly 2 persons out of 5 will recover from the disease?
0.0512
Probability of Recovery: Understanding Binomial Distribution
This question asks for the probability that a specific number of people recover from a disease, given the probability of a single person recovering. This type of problem fits the characteristics of a binomial probability distribution.
What is Binomial Probability?
A binomial distribution applies when we have a fixed number of independent trials, where each trial has only two possible outcomes (often called "success" and "failure"), and the probability of success is the same for every trial.
In this scenario:
- Each person is a trial.
- The number of trials is fixed at 5 persons.
- There are two outcomes for each person: recovery (success) or not recovering (failure).
- The probability of recovery is the same for each person (0.8).
Parameters for the Probability Calculation
Based on the question, we can identify the parameters needed for the binomial probability formula:
- Number of trials, \(n = 5\) (total number of persons).
- Probability of success (a person recovering), \(p = 0.8\).
- Probability of failure (a person not recovering), \(q = 1 - p = 1 - 0.8 = 0.2\).
- Desired number of successes (exactly 2 persons recovering), \(k = 2\).
Binomial Probability Formula
The probability of getting exactly \(k\) successes in \(n\) independent trials is given by the binomial probability formula:
\(P(X=k) = C(n, k) \cdot p^k \cdot q^{n-k}\)
Where \(C(n, k)\) is the binomial coefficient, representing the number of ways to choose \(k\) successes from \(n\) trials. It is calculated as:
\(C(n, k) = \frac{n!}{k!(n-k)!}\)
Calculating the Probability
We want to find the probability that exactly 2 persons out of 5 will recover, which is \(P(X=2)\) with \(n=5\), \(k=2\), \(p=0.8\), and \(q=0.2\).
Step 1: Calculate the binomial coefficient \(C(5, 2)\)
\(C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)} = \frac{120}{2 \times 6} = \frac{120}{12} = 10\)
There are 10 different combinations of selecting exactly 2 persons out of 5 who will recover.
Step 2: Calculate the probability of \(k\) successes (\(p^k\))
The probability of exactly 2 persons recovering is \(p^2\):
\((0.8)^2 = 0.8 \times 0.8 = 0.64\)
Step 3: Calculate the probability of \(n-k\) failures (\(q^{n-k}\))
The number of failures is \(n-k = 5-2 = 3\). The probability of exactly 3 persons not recovering is \(q^3\):
\((0.2)^3 = 0.2 \times 0.2 \times 0.2 = 0.008\)
Step 4: Multiply the results
Now, substitute these values into the binomial probability formula:
\(P(X=2) = C(5, 2) \cdot (0.8)^2 \cdot (0.2)^3\)
\(P(X=2) = 10 \times 0.64 \times 0.008\)
\(P(X=2) = 6.4 \times 0.008\)
\(P(X=2) = 0.0512\)
The probability that exactly 2 persons out of 5 will recover from the disease is 0.0512.
Summary of Calculation
| Parameter | Value | Description |
|---|---|---|
| \(n\) | 5 | Total number of persons |
| \(p\) | 0.8 | Probability of recovery (success) |
| \(q\) | 0.2 | Probability of not recovering (failure) |
| \(k\) | 2 | Exact number of recoveries (successes) |
| \(C(5, 2)\) | 10 | Number of ways to choose 2 from 5 |
| \(p^2\) | 0.64 | Probability of 2 successes |
| \(q^3\) | 0.008 | Probability of 3 failures |
| \(P(X=2)\) | 0.0512 | Final probability |
Checking Options
Comparing our result, 0.0512, with the given options:
- Option 1: 0.00512
- Option 2: 0.02048
- Option 3: 0.2048
- Option 4: 0.0512
Our calculated probability matches Option 4.
Revision Table: Key Concepts in Probability
| Concept | Description | Relevance to Question |
|---|---|---|
| Probability | A measure of the likelihood of an event occurring, usually between 0 and 1. | The question asks for a specific probability. |
| Binomial Distribution | A discrete probability distribution used for situations with a fixed number of independent trials, each having two outcomes with constant probability. | The scenario perfectly fits the criteria for binomial distribution. |
| Independent Trials | The outcome of one trial does not affect the outcome of other trials. | Assumed for each person recovering from the disease. |
| Binomial Coefficient \(C(n, k)\) | The number of ways to choose \(k\) items from a set of \(n\) items without regard to the order. | Needed to account for all possible combinations of exactly 2 people recovering out of 5. |
Additional Information: When to use Binomial Distribution
You should consider using the binomial distribution when your problem involves the following conditions:
- Fixed number of trials (\(n\)): You know exactly how many times the event is repeated.
- Two possible outcomes: Each trial results in either a "success" or a "failure".
- Independent trials: The outcome of one trial does not influence the outcome of any other trial.
- Constant probability of success (\(p\)): The probability of a "success" remains the same for every single trial.
In this problem, we had 5 persons (fixed trials), each could either recover (success) or not (failure), the recovery of one person is independent of others, and the probability of recovery is given as a constant 0.8.
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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:
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