Problem 67 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The mean and variance of a random variable \(X\) having binomial distribution are 4 and 2 respectively, then \(P\) \((X=1)\) is [2003] (a) \(\frac{1}{4}\) (b) \(\frac{1}{32}\) (c) \(\frac{1}{16}\) (d) \(\frac{1}{8}\)

Step-by-Step Solution

Answer

(b) \(\frac{1}{32}\)

Step 1: Understand the Binomial Distribution

The random variable \(X\) follows a binomial distribution, which is denoted as \(X \sim B(n, p)\), where \(n\) is the number of trials and \(p\) is the probability of success in each trial.

Step 2: Relate Mean and Variance to Parameters

For a binomial distribution, the mean \(\mu\) is given by \(\mu = np\) and the variance \(\sigma^2\) is given by \(\sigma^2 = np(1-p)\). We know \(\mu = 4\) and \(\sigma^2 = 2\).

Step 3: Create Equations from Mean and Variance

Use the given mean and variance to form the equations: \(np = 4\) and \(np(1-p) = 2\).

Step 4: Solve for Probability \(p\)

Substitute \(np = 4\) into the variance equation to get \(4(1-p) = 2\). Solve for \(p\): \[4 - 4p = 2 \implies 4p = 2 \implies p = \frac{1}{2}\]

Step 5: Solve for Number of Trials \(n\)

Using \(np = 4\) and \(p = \frac{1}{2}\), solve for \(n\): \[n \times \frac{1}{2} = 4 \implies n = 8\]

Step 6: Calculate \(P(X=1)\)

Use the binomial probability formula: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]Substitute \(n = 8\), \(p = \frac{1}{2}\), and \(k = 1\):\[P(X = 1) = \binom{8}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{7}\]Calculate:\[P(X = 1) = 8 \times \frac{1}{2^8} = \frac{8}{256} = \frac{1}{32}\]

Key Concepts

Probability of Success

The probability of success in a binomial experiment is a crucial concept that determines the likelihood of an event happening in one trial. When we deal with binomial distributions, we denote the probability of success as \(p\). It represents how likely it is for the particular outcome to occur in a single attempt.

In the given exercise, we use the relationship between the mean and variance to find \(p\). We know that the mean (\(\mu\)) of a binomial distribution is calculated as \(\mu = np\), where \(n\) is the number of trials, and \(p\) is the probability of success. Similarly, the variance (\(\sigma^2\)) is expressed as \(np(1-p)\). With these formulas, we have two equations formed from the provided information about the mean and variance.

* Mean Equation: \(np = 4\)
* Variance Equation: \(np(1-p) = 2\)

By solving these equations simultaneously in the solution, we find that \(p = \frac{1}{2}\). This means the probability of success in each trial is 50%. Understanding \(p\) helps us further explore and calculate different probabilities using the binomial distribution.

Mean and Variance

The mean and variance of a binomial distribution are fundamental statistics that summarize the behavior of the random variable. Mean and variance are calculated using the probability of success and number of trials:

- Mean (\(\mu\)): This is given by the formula \(\mu = np\). It represents the expected number of successes over \(n\) trials.

- Variance (\(\sigma^2\)): This is calculated using \(np(1-p)\). The variance tells us how much the number of successes is likely to vary around the mean.

In the example provided:

1. We know the mean is 4, hence \(np = 4\).
2. The variance is given as 2, and using \(np(1-p) = 2\), we relate it to \(p\) and \(n\).

Solving these helps us pinpoint specific characteristics of the distribution, namely the probability \(p\) and the number of trials \(n\). By finding these values and understanding their mathematical relationships, we can predict and understand the distribution’s behavior better.

Binomial Probability Formula

The binomial probability formula is used to find the probability of a specific number of successes in \(n\) independent trials. The formula is expressed as:

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

where:

• \(\binom{n}{k}\) is the binomial coefficient, representing the number of ways to choose \(k\) successes out of \(n\) trials.
• \(p^k\) is the probability of \(k\) successes.
• \((1-p)^{n-k}\) is the probability of \(n-k\) failures.

For our exercise, after deducing \(n = 8\) and \(p = \frac{1}{2}\), we apply the formula to find the probability that \(X = 1\):

\[ P(X = 1) = \binom{8}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^7 \]

Calculating it step-by-step:

• Compute the binomial coefficient: \(\binom{8}{1} = 8\).
• Then, \(\left(\frac{1}{2}\right)^1 = \frac{1}{2}\) and \(\left(\frac{1}{2}\right)^7 = \frac{1}{128}\).

This results in:

\[ P(X = 1) = 8 \times \frac{1}{2^8} = \frac{8}{256} = \frac{1}{32}\]

This calculated probability indicates the chance of success in exactly one trial when conducting eight trials, each with a 50% success rate.