From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let

Q: Wait, why can't I just use the Binomial variance formula $npq$? It's so much faster!

That is a classic trap! The Binomial distribution assumes independent trials with replacement. Here, because we are sampling without replacement, the trials are dependent. Using $npq$ would give you the wrong answer. Always check if the population size is finite and sampling is without replacement—that is your cue for Hypergeometric!

Q: Do I really have to calculate $P(X=0), P(X=1), P(X=2),$ and $P(X=3)$ individually? Isn't there a formula?

For a small sample size like this, calculating them individually is actually safer and less prone to formulaic errors. While there is a general formula for the variance of a hypergeometric distribution, deriving it or remembering it under exam pressure is risky. Stick to the first principles: $E(X)$ and $E(X^2)$. It builds intuition and ensures accuracy.

Q: Why did we reject $X=4$ or $X=5$ as possible values?

Great observation! Look at the constraints of the problem. We only have 3 defective items in the entire lot of 10. Even if we draw 5 items, it is physically impossible to draw 4 or 5 defective items because they simply don't exist in the box. Always define your sample space based on the physical reality of the problem, not just the sample size.

Question

From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $\sigma^2$, then $96\sigma^2$ is equal to

Sigma Percentile · Accepted Answer

### The Art of Counting: Navigating the Hypergeometric Landscape

Welcome, student. Today, we are not just solving a probability problem; we are stepping into the shoes of a quality control engineer. Imagine you are standing before a bin of 10 components.

You know that 3 are defective and 7 are pristine. You are tasked with pulling a sample of 5. This is a scenario of uncertainty where we quantify the risk of finding defective items using the **Hypergeometric distribution**.

### Phase 1: Defining the Sample Space

Before we touch any numbers, we must visualize the total possibilities. We are choosing 5 items out of 10, which is a classic combination problem.

The total number of ways to draw this sample is given by the binomial coefficient $\binom{10}{5}$. Calculating this, we find:

$$
\binom{10}{5} = \frac{10!}{5! 	imes 5!} = 252
$$

This number, $252$, is our universe. Every probability we calculate will be a fraction of this total, serving as the bedrock upon which our distribution rests.

### Phase 2: The Probability Distribution

Let $X$ be the random variable representing the number of defective items in our sample. Since we only have 3 defective items in total, $X$ can only take values from $0$ to $3$.

To find the probability $P(X=r)$, we use the hypergeometric formula:

$$
P(X=r) = \frac{\binom{3}{r} 	imes \binom{7}{5-r}}{\binom{10}{5}}
$$

Think of this as a two-step selection process: first, you choose $r$ defective items from the 3 available, and then you fill the remainder of your sample ($5-r$) from the 7 good items.

When we calculate these for $r=0, 1, 2, 3$, we get a symmetric distribution:

- For $r=0$: $P(X=0) = \frac{\binom{3}{0} 	imes \binom{7}{5}}{252} = \frac{1 	imes 21}{252} = \frac{1}{12}$

- For $r=1$: $P(X=1) = \frac{\binom{3}{1} 	imes \binom{7}{4}}{252} = \frac{3 	imes 35}{252} = \frac{5}{12}$

- For $r=2$: $P(X=2) = \frac{\binom{3}{2} 	imes \binom{7}{3}}{252} = \frac{3 	imes 35}{252} = \frac{5}{12}$

- For $r=3$: $P(X=3) = \frac{\binom{3}{3} 	imes \binom{7}{2}}{252} = \frac{1 	imes 21}{252} = \frac{1}{12}$

### Phase 3: The Statistical Moments

Now, we move to the heart of the problem: the variance, $\sigma^2$. We know the definition of variance is $\sigma^2 = E(X^2) - [E(X)]^2$.

First, let's find the mean, $E(X)$:

$$
E(X) = \sum X \cdot P(X) = 0\left(\frac{1}{12}ight) + 1\left(\frac{5}{12}ight) + 2\left(\frac{5}{12}ight) + 3\left(\frac{1}{12}ight) = \frac{18}{12} = \frac{3}{2}
$$

Next, we calculate the second moment, $E(X^2)$. We square the value of $X$ before multiplying by the probability:

$$
E(X^2) = \sum X^2 \cdot P(X) = 0^2\left(\frac{1}{12}ight) + 1^2\left(\frac{5}{12}ight) + 2^2\left(\frac{5}{12}ight) + 3^2\left(\frac{1}{12}ight) = \frac{34}{12} = \frac{17}{6}
$$

### The Final Act

With $E(X) = \frac{3}{2}$ and $E(X^2) = \frac{17}{6}$, the variance is calculated as follows:

$$
\sigma^2 = \frac{17}{6} - \left(\frac{3}{2}ight)^2 = \frac{17}{6} - \frac{9}{4} = \frac{34 - 27}{12} = \frac{7}{12}
$$

The question asks for the value of $96\sigma^2$. This is our final flourish:

$$
96 	imes \frac{7}{12} = 8 	imes 7 = 56
$$

The final answer is **56**. Through careful counting and systematic application of statistical moments, we have successfully tamed the randomness of the lot.

Animated Solution for Mathematics - Probability: From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $σ^{2}$ , then $96 σ^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Lot

Defining the Sample

Total Possible Outcomes

Probability Distribution Formula

Calculating $P (X = 0)$

Calculating $P (X = 1)$

Calculating $P (X = 2)$

Calculating $P (X = 3)$

Finding the Mean $E (X)$

Calculating $E (X^{2})$

Calculating Variance $σ^{2}$

Final Answer: $96 σ^{2}$

The Sigma Insight: Random Variables and Probability Distributions

The Art of Counting

Phase 1

Phase 2

Phase 3

The Final Act

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