A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the

Q: Wait, why can't we use the binomial distribution formula here?

That is a fantastic question! The binomial distribution assumes that each trial is independent, which happens when we sample *with* replacement. Here, we are drawing three balls simultaneously, which is equivalent to sampling *without* replacement. This changes the probabilities for each subsequent draw, so we must use the hypergeometric approach (combinations) instead.

Q: Why is the variance formula $E(X^2) - [E(X)]^2$ better than $\sum (x_i - \mu)^2 P(x_i)$?

Both formulas are mathematically equivalent! However, $E(X^2) - [E(X)]^2$ is often preferred in competitive exams because it reduces the number of subtraction operations involving the mean, which can sometimes be a non-integer or a complex fraction, thereby minimizing the chance of calculation errors.

Q: How do I know if my probability distribution is correct before calculating variance?

Always perform a quick sanity check: the sum of all probabilities in your distribution must equal exactly 1. If $\sum P(x_i) \neq 1$, you know immediately that you've missed a case or miscalculated a combination. It's the best way to catch errors early!

Question

A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let $X$ be the number of white balls, among the drawn balls. If $\sigma^2$ is the variance of $X$, then $100 \sigma^2$ is equal to

Sigma Percentile · Accepted Answer

### The Dance of Probability: A Journey into Variance

Imagine you are standing before a bag containing ten mysterious spheres—four are white, and six are black. You are tasked with reaching in and pulling out three at once.

We define a random variable $X$ as the number of white balls you hold in your hand. As you prepare to draw, $X$ could be $0, 1, 2,$ or $3$. Our goal is to find the variance of this variable, $\sigma^2$, and ultimately, the value of $100 \sigma^2$.

### Mapping the Sample Space

Before we can predict the future, we must understand the total landscape of possibilities. Since we are selecting three balls out of ten and the order of selection does not matter, we rely on the power of combinations.

The total number of ways to choose these three balls is given by the combination formula:

$$n(S) = {}^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$$

This number, $120$, will serve as the bedrock for all our probability calculations.

### The Probability Distribution

Now, we calculate the probability for each possible value of $X$.

For $X=0$, we draw zero white balls and three black balls. The number of ways is ${}^4C_0 	imes {}^6C_3 = 1 	imes 20 = 20$. Thus:

$$P(X=0) = \frac{20}{120} = \frac{1}{6}$$

For $X=1$, we draw one white ball and two black balls. The number of ways is ${}^4C_1 	imes {}^6C_2 = 4 	imes 15 = 60$. Thus:

$$P(X=1) = \frac{60}{120} = \frac{1}{2}$$

For $X=2$, we draw two white balls and one black ball. The number of ways is ${}^4C_2 	imes {}^6C_1 = 6 	imes 6 = 36$. Thus:

$$P(X=2) = \frac{36}{120} = \frac{3}{10}$$

For $X=3$, we draw three white balls and zero black balls. The number of ways is ${}^4C_3 	imes {}^6C_0 = 4 	imes 1 = 4$. Thus:

$$P(X=3) = \frac{4}{120} = \frac{1}{30}$$

### Expectation and the Second Moment

With our distribution complete, we calculate the expected value, $E(X)$, which represents the long-term average of our experiment:

$$E(X) = \sum x_i P(x_i) = 0\left(\frac{1}{6}ight) + 1\left(\frac{1}{2}ight) + 2\left(\frac{3}{10}ight) + 3\left(\frac{1}{30}ight) = 0 + 0.5 + 0.6 + 0.1 = 1.2$$

Next, we find the second moment, $E(X^2)$, which is crucial for determining the spread of our data:

$$E(X^2) = \sum x_i^2 P(x_i) = 0^2\left(\frac{1}{6}ight) + 1^2\left(\frac{1}{2}ight) + 2^2\left(\frac{3}{10}ight) + 3^2\left(\frac{1}{30}ight) = 0 + 0.5 + 1.2 + 0.3 = 2.0$$

### The Final Calculation

We are now at the climax of our mathematical journey. The variance $\sigma^2$ is defined as the difference between the second moment and the square of the expectation:

$$\sigma^2 = E(X^2) - [E(X)]^2$$

Substituting our values, we get:

$$\sigma^2 = 2.0 - (1.2)^2 = 2.0 - 1.44 = 0.56$$

The question asks for $100 \sigma^2$, so we multiply our result by $100$ to arrive at the final, satisfying integer:

$$100 	imes 0.56 = \mathbf{56}$$

Animated Solution for Mathematics - Probability: A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let $X$ be the number of white balls, among the drawn balls. If $σ^{2}$ is the variance of $X$ , then $100 σ^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Total Sample Space $n (S)$

Probability of $X = 0$

Probability of $X = 1$

Probability of $X = 2$

Probability of $X = 3$

Calculating Expectation $E (X)$

Calculating $E (X^{2})$

Finding Variance $σ^{2}$

Final Answer: $100 σ^{2}$

The Sigma Insight: Random Variables and Probability Distributions

The Dance of Probability

Mapping the Sample Space

The Probability Distribution

Expectation and the Second Moment

The Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Total Sample Space .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(S)

Probability of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=0

Probability of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=1

Probability of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=2

Probability of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=3

Calculating Expectation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E(X)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E(X2)

Finding Variance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ2

Final Answer: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 100σ2