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

First, let us recognise that drawing “with replacement” means the composition of the bag never changes. Therefore, for every single draw, the probability of picking a white ball remains constant.

The bag has 30 white balls and 10 red balls, altogether 40 balls. So the probability of getting a white ball in any one draw is

$$p=\frac{\text{number of white balls}}{\text{total balls}}=\frac{30}{40}=\frac34.$$

Because each of the 16 draws is independent and has the same probability of success (drawing a white ball), the random variable $$X=$$ “number of white balls obtained in 16 draws” follows a binomial distribution. We write this as

$$X\sim\text{Binomial}(n=16,\;p=\tfrac34).$$

For a binomially distributed variable we use these standard formulae:   • Mean (expected value) $$\mu = np.$$   • Variance $$\sigma^{2} = npq,$$ where $$q=1-p.$$   • Standard deviation $$\sigma = \sqrt{npq}.$$

We substitute $$n=16$$ and $$p=\tfrac34$$.

First, find the mean:

$$\mu = np = 16\left(\frac34\right)=\frac{16\times3}{4}=12.$$

Next, compute $$q$$:

$$q = 1-p = 1-\frac34 = \frac14.$$

Now, find the variance:

$$\sigma^{2} = npq = 16\left(\frac34\right)\left(\frac14\right) = 16 \times \frac{3}{16} = 3.$$

The standard deviation is the square root of the variance, so

$$\sigma = \sqrt{3}.$$

The question asks for the ratio of the mean of $$X$$ to the standard deviation of $$X$$. We therefore write

$$\frac{\text{mean of }X}{\text{standard deviation of }X}=\frac{\mu}{\sigma}= \frac{12}{\sqrt{3}}.$$

To simplify $$\frac{12}{\sqrt{3}}$$, we multiply numerator and denominator by $$\sqrt{3}$$:

$$\frac{12}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{12\sqrt{3}}{3}=4\sqrt{3}.$$

Hence, the correct answer is Option B.