Problem 57 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the mean and the variance of a binomial variate \(\mathrm{X}\) are 2 and 1 respectively, then the probability that \(X\) takes a value greater than or equal to one is : [Online April 11, 2015] (a) \(\frac{9}{16}\) (b) \(\frac{3}{4}\) (c) \(\frac{1}{16}\) (d) \(\frac{15}{16}\)

Step-by-Step Solution

Answer

The probability that \(X\) takes a value greater than or equal to one is \(\frac{15}{16}\) (option d).

Step 1: Understand Binomial Distribution Parameters

A binomial variate \(X\) has parameters \(n\) (number of trials) and \(p\) (probability of success). The mean \( \mu = np \) and the variance \( \sigma^2 = np(1-p) \). In this problem, we are given that \( \mu = 2 \) and \( \sigma^2 = 1 \).

Step 2: Set Up Equations for Mean and Variance

We have two equations: \( np = 2 \) and \( np(1-p) = 1 \). These equations relate the parameters \(n\) and \(p\) of the binomial distribution.

Step 3: Solve for Probability of Success \(p\)

Substitute \( np = 2 \) into the equation \( np(1-p) = 1 \). We get \( 2(1-p) = 1 \). Solve this equation to find \(p\).\[ 2 - 2p = 1 \] \[ 2p = 1 \] \[ p = \frac{1}{2} \]

Step 4: Solve for Number of Trials \(n\)

Using \(np = 2\) and \(p = \frac{1}{2}\), substitute \(p\) back to find \(n\).\[ n \left( \frac{1}{2} \right) = 2 \] \[ n = 4 \]

Step 5: Calculate P(X ≥ 1)

The probability that \(X\) is greater than or equal to 1 is \(1 - P(X = 0)\). For \(X = 0\), substitute \(n = 4\) and \(p = \frac{1}{2}\) into the binomial probability formula: \(P(X = 0) = \binom{4}{0} \left( \frac{1}{2} \right)^0 \left( \frac{1}{2} \right)^4\). \(P(X \, = \, 0) = \left( \frac{1}{2} \right)^4 = \frac{1}{16}\). Therefore, \(P(X \, \geq \, 1) = 1 - \frac{1}{16} = \frac{15}{16}\).

Key Concepts

Probability of Success

In a binomial distribution, the probability of success, denoted as \( p \), is the chance of a single trial resulting in a success. Understanding this probability is crucial to solving problems involving binomial distributions. Let's say we are conducting an experiment that can result in either a success or failure, like tossing a coin. The probability of success in each trial will directly impact the overall results we calculate.
In the original exercise, we were given that the mean \( \mu \) of a binomial distribution is 2, and the variance \( \sigma^2 \) is 1. These values relate to the probability of success \( p \) and the number of trials \( n \). We use the relationships:

• Mean: \( \mu = np \)
• Variance: \( \sigma^2 = np(1-p) \)

By substituting \( np = 2 \) into \( np(1-p) = 1 \) and solving, we find that the probability of success \( p \) is \( \frac{1}{2} \).
This small number may seem simple, but it plays a huge role in telling us the likelihood of success in any given trial within the distribution.

Mean and Variance

The mean and variance of a binomial distribution offer insightful information about the distribution's characteristics. The mean, \( \mu \), is essentially the expected number of successes in \( n \) trials. It's calculated as \( \mu = np \), where \( n \) is the number of trials, and \( p \) is the probability of success.
Variance, shown as \( \sigma^2 \), measures the dispersion or spread of the distribution. In a binomial distribution, it's calculated as \( \sigma^2 = np(1-p) \). A lower variance indicates that the results are tightly clustered around the mean, while a higher variance suggests a wider spread of results.
In the given exercise, the mean \( \mu \) was 2, and the variance \( \sigma^2 \) was 1. These numbers allowed us to derive both the probability of success and the number of trials:

• From \( np = 2 \), we used \( p = \frac{1}{2} \) to calculate the number of trials \( n = 4 \).

This connection between mean and variance is critical in fully understanding the nature of a binomial distribution and predicting outcomes.

Probability Calculation

Calculating probabilities in a binomial distribution often revolves around determining specific outcome probabilities through the binomial formula. The probability of any specific outcome \( X = k \) in our distribution is given by:

• \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)

Here, \( \binom{n}{k} \) is a combination representing the number of ways to choose \( k \) successes from \( n \) trials.
In the original exercise, we aimed to calculate the probability that \( X \) is greater than or equal to 1, i.e., \( P(X \geq 1) \). As a shortcut, we may calculate the complement \( 1 - P(X = 0) \). Therefore, we analyzed the likelihood of no successes \( X = 0 \) and found:

• \( P(X = 0) = \binom{4}{0} \left( \frac{1}{2} \right)^0 \left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16} \)

This resulted in \( P(X \geq 1) = 1 - \frac{1}{16} = \frac{15}{16} \), revealing a high probability of achieving at least one success in the trials.
Understanding these calculations is vital for anyone working with binomial probabilities as they guide predictions and decisions.