A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one–by–one, with replac

Q: Wait, why do we use Binomial Distribution here? Can't we use Hypergeometric?

That is a brilliant question! We use the Binomial Distribution specifically because of the phrase 'with replacement'. If we were not replacing the balls, the probability of picking a green ball would change after every draw, forcing us into the Hypergeometric realm. Because we replace them, each draw is an independent event with a constant probability, which is the definition of a Bernoulli process.

Q: Why is the variance $npq$ and not just $np$?

Great catch! $np$ is the mean (the expected value) of the distribution—it tells you where the 'center' of your results lies. The variance, $npq$, measures the 'spread' or the uncertainty around that mean. Since $q$ is always less than 1, the variance is always smaller than the mean in a Binomial distribution.

Q: Does the order of the balls matter in this calculation?

In the context of the Binomial distribution, we only care about the *total count* of green balls, not the sequence in which they appear. The Binomial model inherently accounts for all possible combinations of successes and failures, so you don't need to worry about the order!

Q: What if I accidentally used $q = 15/25$? How would that affect the answer?

If you swapped $p$ and $q$, you would be calculating the variance of the number of *yellow* balls instead of green balls. Interestingly, because the variance formula $npq$ is symmetric with respect to $p$ and $q$, you would get the same numerical result! However, it is vital to keep your definitions consistent to avoid confusion in more complex problems.

Q: Is there a shortcut to remember the variance formula?

Think of it as the 'spread' of a coin toss. For a single coin flip, the variance is $pq$. When you perform $n$ independent flips, the variances simply add up, giving you $n \cdot p \cdot q$. It is a beautiful, additive property of independent random variables!

Question

A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is:

Options:

(A) $\frac{12}{5}$

(B) $6$

(C) $4$

(D) $\frac{6}{25}$

Sigma Percentile · Accepted Answer

### The Dance of Probability: Understanding Variance

Imagine you are standing before a box filled with $15$ green balls and $10$ yellow balls. You are tasked with drawing $10$ balls, one by one.

Every time you pull a ball out, you look at its color, record it, and then—crucially—you drop it back into the box. This simple act of 'replacement' transforms a complex, dependent sequence into a series of independent, identical trials. This is the heart of the **Binomial Distribution**.

### Defining the Landscape

First, let us look at our total population. With $15$ green and $10$ yellow, we have a total of $25$ balls.

The probability of success—drawing a green ball—is defined as:

$$p = \frac{15}{25} = \frac{3}{5}$$

Consequently, the probability of failure (drawing a yellow ball) is:

$$q = 1 - p = \frac{10}{25} = \frac{2}{5}$$

Because we replace the ball each time, these probabilities remain constant. We are performing $n = 10$ independent trials.

### The Soul of the Variance

In the world of statistics, when we talk about the Binomial Distribution $X \sim B(n, p)$, we are interested in how much our results fluctuate. The variance $\sigma^2$ quantifies the 'spread' of our data.

The formula is elegant and powerful:

$$\sigma^2 = npq$$

This formula represents the accumulation of uncertainty across $n$ independent events. Each event has a variance of $pq$, and because they are independent, we simply multiply by $n$.

### The Calculation

Now, let us bring our numbers into the fold. We have $n = 10$, $p = \frac{3}{5}$, and $q = \frac{2}{5}$. Substituting these into our variance equation, we get:

$$\sigma^2 = 10 \cdot \left(\frac{3}{5}\right) \cdot \left(\frac{2}{5}\right)$$

Let us tackle the product of the probabilities first. Multiplying $\frac{3}{5}$ and $\frac{2}{5}$ gives us $\frac{6}{25}$.

Now, we multiply this by our number of trials, $n = 10$:

$$\sigma^2 = 10 \cdot \frac{6}{25} = \frac{60}{25}$$

### The Final Simplification

To reach the final, cleanest form of our answer, we look at the fraction $\frac{60}{25}$. Both the numerator and the denominator share a common factor of $5$.

Dividing both by $5$, we arrive at:

$$\sigma^2 = \frac{12}{5}$$

The variance of the number of green balls drawn is **$\frac{12}{5}$**. This value is not just a number; it is a measure of the inherent randomness in your experiment. By understanding the conditions—the 'with replacement' clause—you have successfully navigated the Binomial Distribution.

Animated Solution for Mathematics - Probability: A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is:

Select Answer:

Visualized Solution

The Setup

Total Number of Balls

Identifying the Distribution

Number of Trials ( $n$ )

Probability of Success ( $p$ )

Probability of Failure ( $q$ )

Variance of Binomial Distribution

Substituting the Values

Multiplying the Probabilities

Final Multiplication

Final Simplification

The Sigma Insight: Random Variables and Probability Distributions

The Dance of Probability

Defining the Landscape

The Soul of the Variance

The Calculation

The Final Simplification

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