Problem 156 - A Complete Resource Book in Mathematics for JEE Main

Question

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is (A) \(\frac{37}{256}\) (B) \(\frac{219}{256}\) (C) \(\frac{128}{256}\) (D) \(\frac{28}{256}\)

Step-by-Step Solution

Answer

Option (D) \( \frac{28}{256} \).

Step 1: Identify Binomial Distribution Parameters

Given that the mean of a binomial distribution is 4 and the variance is 2, we can connect these to the parameters of the binomial distribution. The mean of a binomial distribution is given by \( np \) and the variance by \( np(1-p) \). Using these, we can set up two equations: \( np = 4 \) and \( np(1-p) = 2 \).

Step 2: Solve for n and p

Let's solve the equations to find \( n \) and \( p \). From \( np = 4 \), \( n = \frac{4}{p} \). Substitute this into the variance equation: \( \frac{4}{p}p(1-p) = 2 \). Simplifying gives: \( 4(1-p) = 2 \). So, \( 4 - 4p = 2 \), thus \( p = \frac{1}{2} \) and \( n = 8 \).

Step 3: Substitute Values in Probability Formula

With \( n = 8 \), \( p = \frac{1}{2} \), the probability mass function for a binomial distribution is \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). We want \( P(X = 2) = \binom{8}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{8-2}\).

Step 4: Calculate the Probability

Calculate \( \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \). Then compute the probability: \( P(X = 2) = 28 \left(\frac{1}{2}\right)^8 = \frac{28}{256} \).

Step 5: Choose the Correct Option

The calculated probability \( \frac{28}{256} \) matches option (D). Therefore, the probability of 2 successes is (D) \( \frac{28}{256} \).

Key Concepts

Mean of Binomial Distribution

The mean of the binomial distribution is a central concept and can be thought of as the average number of successes we would expect if we repeat the experiment a large number of times. In mathematical terms, the mean, or expected value, of a binomial distribution is given by the formula:\[ \mu = np \]where:- \( \mu \) is the mean.- \( n \) is the number of trials or the sample size.- \( p \) is the probability of success in each trial.For the distribution in our exercise, the mean is given as 4. This directs us to the equation \( np = 4 \). To understand the intuition behind this, imagine flipping a coin with \( n \) total flips and probability \( p \) for landing heads. The mean would be how many heads we expect to see on average. The mean helps to give us a sense of the central tendency or typical outcome in a binomial distribution.

Variance of Binomial Distribution

The variance of a binomial distribution helps describe how spread out the successes are around the mean. It is a measure of the dispersion or variability of the outcomes.Mathematically, the variance is expressed as:\[ \sigma^2 = np(1-p) \]where:- \( \sigma^2 \) is the variance.- \( n \) and \( p \) are the same as described earlier.- \( 1-p \) represents the probability of failure.In this instance, the variance is given as 2. So, using our understanding of the relationships between the mean, variance, and the probabilities, i.e., the equations \( np = 4 \) and \( np(1-p) = 2 \), we can solve for \( p \) and subsequently \( n \) to ensure these values make sense practically. Variance indicates how much the number of successes varies from the mean number of expected successes. A higher variance means greater variability in the number of successes per set of trials.

Probability Mass Function

The probability mass function (PMF) of a binomial distribution gives the probability of obtaining a specific number of successes in a fixed number of trials, each with the same probability of success. It is fundamental for calculating the likelihood of different outcomes.The PMF of a binomial distribution is expressed as:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where:- \( P(X = k) \) is the probability of observing \( k \) successes.- \( \binom{n}{k} \) is a binomial coefficient calculated as \( \frac{n!}{k!(n-k)!} \).- \( p^k \) is the probability of success raised to the number of successes.- \( (1-p)^{n-k} \) accounts for the probability of the remaining trials being failures.In our exercise, we calculated the probability of exactly 2 successes when \( n = 8 \) and \( p = \frac{1}{2} \) using this function. The PMF allows us to determine the probability of various potential outcomes, making it a powerful tool for understanding and predicting results within the framework of a binomial distribution. It encapsulates the essence of evaluating different possible results given a certain configuration of a binomial experiment.