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

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

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.