Problem 54 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If \(X\) be the number of white balls drawn, then \(\left(\frac{\text { mean of } \mathrm{X}}{\text { standard deviation of } \mathrm{X}}\right)\) is equal to:[Jan. 11, 2019 (II)] (a) 4 (b) \(4 \sqrt{3}\) (c) \(3 \sqrt{2}\) (d) \(\frac{4 \sqrt{3}}{3}\)

Step-by-Step Solution

Answer

The answer is (a) 4.

Step 1: Calculate the probability of drawing a white ball

First, find the probability of drawing a white ball from the bag. There are 30 white balls out of a total of 30 + 10 = 40 balls. So, the probability \(P(W)\) of drawing a white ball is \(P(W) = \frac{30}{40} = \frac{3}{4}\).

Step 2: Determine the mean of X

Since each draw is independent and with replacement, the random variable \(X\), representing the number of white balls drawn out of 16, follows a binomial distribution. The mean \(\mu\) of a binomial distribution \(X \sim B(n, p)\) is given by \(\mu = n \times p\). Here, \(n = 16\) and \(p = \frac{3}{4}\). Thus, \(\mu = 16 \times \frac{3}{4} = 12\).

Step 3: Determine the variance and standard deviation of X

For a binomial distribution, the variance \(\sigma^2\) is given by \(\sigma^2 = n \times p \times (1-p)\). Substituting, we have \(\sigma^2 = 16 \times \frac{3}{4} \times \frac{1}{4} = 3\). The standard deviation \(\sigma\) is the square root of the variance: \(\sigma = \sqrt{3}\).

Step 4: Calculate the ratio of the mean to the standard deviation

Now, compute \(\frac{\text{mean of } X}{\text{standard deviation of } X}\). From previous calculations, we have \(\frac{12}{\sqrt{3}} = 4 \times \frac{\sqrt{3}}{\sqrt{3}} = 4\).

Key Concepts

Binomial Distribution

The binomial distribution is a fundamental concept in probability theory that applies when we perform a series of independent experiments, or trials, under the same conditions. Each trial has exactly two outcomes, which we usually call "success" and "failure". If we let the random variable \(X\) denote the number of successes in \(n\) trials, and \(p\) represent the probability of success for each trial, then \(X\) follows a binomial distribution. This is often denoted as \(X \sim B(n, p)\).
In the context of the exercise, drawing a white ball can be considered a "success", and since each draw is independent and we replace the ball after drawing, the trials are identical and independent. We draw 16 times, so \(n = 16\), and since we've calculated the probability of drawing a white ball as \(\frac{3}{4}\), we have \(p = \frac{3}{4}\). This setup allows us to use the binomial distribution to analyze the number of white balls drawn.

Mean and Variance

In probability, two important statistical measures for a random variable following a binomial distribution are the mean (expected value) and variance. The mean \(\mu\) gives us a measure of central tendency, essentially indicating the average number of successes in \(n\) trials if we repeated the experiment an infinite number of times. For a binomial distribution, the mean is calculated as \(\mu = n \times p\). In our specific problem, the mean number of white balls drawn is calculated as \(16 \times \frac{3}{4} = 12\).
Variance, on the other hand, shows us how much the outcomes vary around the mean. It is calculated for a binomial distribution as \(\sigma^2 = n \times p \times (1-p)\). Here, \(\sigma^2 = 16 \times \frac{3}{4} \times \left(1 - \frac{3}{4}\right) = 3\). The standard deviation \(\sigma\), which is the square root of the variance, thus \(\sigma = \sqrt{3}\), provides a more intuitive sense of the spread of the data.

Random Variables

A random variable is a numerical description of the outcome of an experiment. It assigns a numerical value to each event in the sample space of a probabilistic experiment. Random variables can be discrete or continuous. In the context of this exercise, we consider a discrete random variable \(X\), representing the number of white balls drawn from the bag.
Since drawing each ball from the bag with replacement fits into defined categories (white or not white), this points to a discrete random variable scenario, where our outcomes are distinct (count of white balls drawn). Each draw has its own outcome probability, dictated by the underlying probabilities of the event - in this case, \(P(W) = \frac{3}{4}\).
The significance of understanding random variables surrounds their ability to map outcomes into a mathematical context, allowing us to then apply probabilistic and statistical tools to assess and interpret the results effectively.