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, the $\left(\frac{\text{mean} \; \text{of} \; X}{\text{standard} \; \text{deviation} \; \text{of} \; X }\right)$ is equal to :

• A. $4$
• B. $\frac{4 \sqrt{3}}{3}$
• C. $ 4 \sqrt{3} $
• D. $ 3 \sqrt{2} $

Answer 1

The Correct Option is C

1. To solve this problem, we first define the variables and understand what is being asked. We have a bag containing a total of \(40\) balls (\(30\) white balls and \(10\) red balls). We draw \(16\) balls randomly with replacement, and \(X\) represents the number of white balls drawn.
2. Since the balls are drawn with replacement, each draw is an independent Bernoulli trial with two possible outcomes: drawing a white ball or a red ball.
3. The probability of drawing a white ball in a single trial is \(P(\text{White}) = \frac{30}{40} = \frac{3}{4}\).
4. Given this setup, \(X\), the number of white balls drawn in \(16\) trials, follows a Binomial distribution with parameters \(n = 16\) (number of trials) and \(p = \frac{3}{4}\) (probability of success).
5. For a Binomial random variable \(X\), the mean \(\mu\) is given by: \(\mu = n \cdot p = 16 \cdot \frac{3}{4} = 12\).
6. The standard deviation \(\sigma\) is given by: \(\sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{16 \cdot \frac{3}{4} \cdot \frac{1}{4}} = \sqrt{3}\).
7. Now, we need to compute the ratio \(\frac{\text{mean of } X}{\text{standard deviation of } X}\):
   • \(\frac{\mu}{\sigma} = \frac{12}{\sqrt{3}}\)
   • Rationalizing the denominator, we multiply by \(\frac{\sqrt{3}}{\sqrt{3}}\): \(\frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}\).
8. Thus, the required value \(\frac{\mu}{\sigma}\) is \(4\sqrt{3}\), which corresponds to the correct answer:
9. \(4\sqrt{3}\) (Option C).