Problem 7 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

The mean and variance of a binomial distribution are 4 and 3, respectively, then the probability of getting exactly 6 successes in this distribution is (a) \({ }^{16} C_{6}\left(\frac{1}{4}\right)^{10}\left(\frac{3}{4}\right)^{6}\) (b) \({ }^{16} C_{6}\left(\frac{1}{4}\right)^{6}\left(\frac{3}{4}\right)^{10}\) (c) \({ }^{12} C_{6}\left(\frac{1}{4}\right)^{10}\left(\frac{3}{4}\right)^{6}\) (d) \({ }^{12} C_{6}\left(\frac{1}{4}\right)^{6}\left(\frac{3}{4}\right)^{6}\)

Step-by-Step Solution

Answer

Option (b) is correct: \({ }^{16} C_{6} \left(\frac{1}{4}\right)^{6} \left(\frac{3}{4}\right)^{10}\).

Step 1: Identify Parameters

The mean of a binomial distribution is given by the formula \(np = 4\), and the variance is given by \(np(1-p) = 3\). Here, \(n\) is the number of trials and \(p\) is the probability of success on each trial.

Step 2: Solve for Probability of Success

We have two equations: \(np = 4\) and \(np(1-p) = 3\). Divide the second equation by the first: \(1-p = \frac{3}{4}\), which means \(p = \frac{1}{4}\).

Step 3: Solve for Number of Trials

Substitute \(p = \frac{1}{4}\) into \(np = 4\) to get \(n \times \frac{1}{4} = 4\). Solving this, we find \(n = 16\).

Step 4: Apply Binomial Probability Formula

The probability of getting exactly 6 successes is given by the binomial probability formula: \[P(X = 6) = { }^{16} C_{6} \left(\frac{1}{4}\right)^6 \left(\frac{3}{4}\right)^{10}\]

Step 5: Identify Correct Answer

Match the obtained expression with the given options. The expression \({ }^{16} C_{6} \left(\frac{1}{4}\right)^6 \left(\frac{3}{4}\right)^{10}\) corresponds to option (b).

Key Concepts

Binomial Probability Formula

The binomial probability formula is a cornerstone in probability theory, particularly useful for calculating the likelihood of a specific number of successes in a fixed number of independent trials. Each trial must have only two possible outcomes: success or failure. This formula is expressed as:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]In the formula:

• \( \binom{n}{k} \) represents the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \), indicating how many ways \( k \) successes can occur in \( n \) trials.
• \( p \) is the probability of success in a single trial.
• \( (1-p) \) is the probability of failure.
• \( n \) is the total number of trials.
• \( k \) is the number of successes you want to find the probability for.

This formula allows us to calculate the probability of exactly \( k \) successes in \( n \) trials, considering each trial is independent. In the given exercise, using \( n = 16 \), \( k = 6 \), \( p = \frac{1}{4} \), and \( 1-p = \frac{3}{4} \), we find the probability that exactly 6 out of 16 trials are successful.

Mean of Binomial Distribution

The mean, or expected value, of a binomial distribution gives us an estimate of the average number of successes over a large number of trials. It is a succinct way to understand the central tendency of a binomial experiment and is computed as:\[ \mu = np \]Here, \( \mu \) is the mean, \( n \) is the number of trials, and \( p \) is the probability of success on each individual trial. This formula indicates that the mean is a function of both the number of trials and the likelihood of success in each trial. In our exercise, the calculated mean is 4, representing the expected number of successes in 16 trials when the probability of success per trial is 0.25. This helps us determine the typical outcome we might expect if the trial were repeated many times.

Variance of Binomial Distribution

Variance is a measure of how spread out the outcomes of a binomial distribution are. It tells us how much the number of successes is expected to vary from the mean. The variance of a binomial distribution is given by:\[ \sigma^2 = np(1-p) \]Where:

• \( \sigma^2 \) represents variance.
• \( n \) is the number of trials.
• \( p \) is the probability of success.
• \( 1-p \) is the probability of failure.

In the context of our exercise, the variance is found to be 3, indicating the spread of the distribution around the mean. This spread is caused by the inherent randomness of each trial outcome and the probability distribution of successes.

Probability of Success

The probability of success in a binomial distribution is a crucial parameter as it directly influences both the mean and variance of the distribution. It is denoted by \( p \) and reflects the chance that a single trial results in success.In our problem, two foundational equations are used to derive \( p \):

• \( np = 4 \) represents the equation for the mean.
• \( np(1-p) = 3 \) represents the equation for the variance.

By dividing the variance equation by the mean equation, we obtain: \( 1-p = \frac{3}{4} \), leading to \( p = \frac{1}{4} \). This value of \( p \) suggests that in each trial, there's a 25% chance of success. This fundamental probability influences the overall distribution properties, such as predicting how many successes are likely and how much variation exists around this number over a series of trials.