A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag

Q: Wait, why do we use the Binomial Distribution here? Can't we just use basic probability?

That is a great question! You *could* think about it intuitively, but the Binomial Distribution is the formal, robust mathematical framework designed exactly for this scenario. Because we have a fixed number of trials ($n=16$) and each trial is independent (thanks to the 'with replacement' condition), the Binomial model is the most efficient and error-proof way to handle the problem. It saves you from getting lost in complex combinatorics.

Q: What if the question said 'without replacement'? Would the answer change?

Absolutely! If we were drawing without replacement, the probability of drawing a white ball would change after every single draw. That would transform the problem into a Hypergeometric Distribution. Always check the 'replacement' condition first—it is the gatekeeper of which statistical model you choose.

Q: Is there a shortcut for the ratio $\frac{\mu}{\sigma}$?

Yes, there is! If you look at the formulas, $\frac{\mu}{\sigma} = \frac{np}{\sqrt{npq}} = \sqrt{\frac{np}{q}}$. You can use this derived formula to save time, but in the JEE, I always recommend calculating $\mu$ and $\sigma$ separately first. It helps you catch silly calculation errors before they ruin your final answer.

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 $\left(\frac{	ext{mean of } X}{	ext{Standard deviation of } X}ight)$ is equal to:

Options:

(A) $3\sqrt{2}$

(B) 4

(C) $\frac{4\sqrt{3}}{3}$

(D) $4\sqrt{3}$

Sigma Percentile · Accepted Answer

### The Bag of Mystery: A Journey into Probability

Imagine you are standing in a quiet room. In front of you sits a simple, opaque bag containing $30$ white balls and $10$ red balls.

Your task is to pull out a ball, record its color, and—most critically—replace it. You repeat this process $16$ times. This is a classic JEE Advanced problem that tests your ability to recognize the underlying structure of randomness.

### Phase 1: The Anatomy of the Problem

The phrase "with replacement" is the heartbeat of this problem. In probability, replacing the ball resets the universe, ensuring the probability of drawing a white ball remains constant for every trial.

Because each draw is independent and has only two possible outcomes—success (white) or failure (red)—we are firmly in the territory of the **Binomial Distribution**. We define our parameters as follows:

$$n = 16$$

$$p = \frac{30}{40} = \frac{3}{4}$$

$$q = 1 - p = \frac{1}{4}$$

### Phase 2: The Statistical Engine

Now that we have identified the Binomial Distribution, we utilize its standard properties. The mean, or the expected value $\mu$, is given by the formula $\mu = np$.

If we perform $16$ trials with a $75\%$ chance of success, we calculate the expected number of white balls:

$$\mu = 16 	imes \frac{3}{4} = 12$$

Next, we calculate the variance $V = npq$, which measures the "spread" or uncertainty of our results. Substituting our values:

$$V = 16 	imes \frac{3}{4} 	imes \frac{1}{4} = 3$$

Since the standard deviation $\sigma$ is the square root of the variance, we find:

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

### Phase 3: The Final Synthesis

We now calculate the ratio of the mean to the standard deviation, $\frac{\mu}{\sigma}$. Given $\mu = 12$ and $\sigma = \sqrt{3}$, the expression is:

$$\frac{\mu}{\sigma} = \frac{12}{\sqrt{3}}$$

To simplify this without a calculator, we express $12$ as $4 	imes \sqrt{3} 	imes \sqrt{3}$:

$$\frac{4 	imes \sqrt{3} 	imes \sqrt{3}}{\sqrt{3}} = 4\sqrt{3}$$

The final result is **$4\sqrt{3}$**.

### The JEE Mindset

The JEE Advanced is not just about finding the answer; it is about recognizing the pattern. By identifying the Binomial Distribution early, we bypassed the need for complex summation or tedious counting.

Whenever you encounter "independent trials" and "two outcomes," let the Binomial Distribution be your first thought. Keep practicing, and you will find that even the most complex problems start to feel like old friends. You have got this!

Animated Solution for Mathematics - Probability: 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 $(\frac{mean of X}{Standard deviation of X})$ is equal to:

Select Answer:

Visualized Solution

Analyzing the Bag Contents

Identifying the Distribution

Defining Parameters $n$ , $p$ , and $q$

Calculating the Mean $μ$

Calculating Variance $V$

Calculating Standard Deviation $σ$

Setting up the Ratio $\frac{μ}{σ}$

Simplifying the Ratio

Final Conclusion

The Sigma Insight: Binomial Distribution

The Bag of Mystery

Phase 1

Phase 2

Phase 3

The JEE Mindset

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