A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X den

Q: Wait, why can't I just use the Binomial Distribution formula here? Isn't this just a series of Bernoulli trials?

That is a fantastic question, and it is a trap many students fall into! The Binomial Distribution requires independent trials, which implies sampling *with* replacement. Here, we are drawing 2 pens from a box of 10. Once you pick the first pen, the composition of the box changes for the second pick. This is sampling *without* replacement, which follows a Hypergeometric Distribution. Always check if the probability changes after each event!

Q: Why do we calculate $E(X^2)$ separately? Can't I just square the expectation $E(X)$?

Absolutely not! That is a common algebraic misconception. $E(X^2)$ is the average of the squares of the outcomes, while $[E(X)]^2$ is the square of the average outcome. They are fundamentally different quantities. The variance is defined as the difference between these two: $Var(X) = E(X^2) - [E(X)]^2$. If they were the same, the variance of every random variable would be zero, which would make statistics very boring!

Q: Is there a faster way to solve this without calculating every single probability?

For this specific problem with only three possible outcomes ($0, 1, 2$), the direct method is actually the most robust and least prone to error. While there are formulas for the variance of a Hypergeometric distribution, deriving them during an exam takes more time and mental energy than simply building the distribution table. Stick to the table method; it is elegant, logical, and keeps you grounded in the physics of the problem.

Question

A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let $X$ denote the number of defective pens. Then the variance of $X$ is

Options:

(A) $\frac{11}{15}$

(B) $\frac{28}{75}$

(C) $\frac{2}{15}$

(D) $\frac{3}{5}$

Sigma Percentile · Accepted Answer

### The Geometry of Uncertainty: Mastering the Variance of Random Variables

Welcome, future engineers. Today, we are not just solving a probability problem; we are learning to quantify the 'risk' or 'uncertainty' inherent in a system.

Imagine you are a quality control engineer at a manufacturing plant. You have a box of $10$ pens, and you know that $3$ are defective. You need to assess the quality of the batch by drawing a sample of $2$.

This is the essence of statistical inference. We are going to walk through this step-by-step, not just to find the answer, but to understand the soul of the math.

### Phase 1: Defining the Universe

Every probability problem begins by defining the sample space. We have a total of $10$ pens and we are choosing $2$.

The total number of ways to do this is given by the combination formula $^{10}C_2$. Mathematically, this is:

$$
\frac{10 	imes 9}{2 	imes 1} = 45
$$

This number, $45$, is our universe. Every probability we calculate will be a fraction of this total.

Our random variable $X$ is the number of defective pens. Since we are drawing $2$ pens, $X$ can be $0$ (no defectives), $1$ (one defective), or $2$ (both defective).

### Phase 2: Mapping the Probabilities

Now, we calculate the likelihood of each scenario.

For $X=0$, we need to choose $0$ defective pens from $3$ and $2$ good pens from $7$:

$$
P(X=0) = \frac{^3C_0 	imes ^7C_2}{45} = \frac{1 	imes 21}{45} = \frac{7}{15}
$$

For $X=1$, we choose $1$ defective from $3$ and $1$ good from $7$:

$$
P(X=1) = \frac{^3C_1 	imes ^7C_1}{45} = \frac{3 	imes 7}{45} = \frac{21}{45} = \frac{7}{15}
$$

Finally, for $X=2$, we choose $2$ defective pens from $3$:

$$
P(X=2) = \frac{^3C_2}{45} = \frac{3}{45} = \frac{1}{15}
$$

Notice how the probabilities sum to $1$: $\frac{7}{15} + \frac{7}{15} + \frac{1}{15} = \frac{15}{15} = 1$. This is our sanity check; if they didn't sum to $1$, we would know we made a mistake.

### Phase 3: The Expectation

The expectation, $E(X)$, is the weighted average of all possible outcomes. It tells us what to expect on average if we repeated this experiment thousands of times.

We calculate it as $E(X) = \sum x_i P(x_i)$. Plugging in our values:

$$
E(X) = 0\left(\frac{7}{15}ight) + 1\left(\frac{7}{15}ight) + 2\left(\frac{1}{15}ight) = 0 + \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5}
$$

This means, on average, we expect $0.6$ defective pens in our sample.

### Phase 4: The Variance

Now, we arrive at the heart of the problem: the variance. Variance measures the spread of the data.

To find it, we first need the second moment, $E(X^2)$. We square the outcomes before multiplying by their probabilities:

$$
E(X^2) = 0^2\left(\frac{7}{15}ight) + 1^2\left(\frac{7}{15}ight) + 2^2\left(\frac{1}{15}ight) = 0 + \frac{7}{15} + \frac{4}{15} = \frac{11}{15}
$$

Finally, we use the variance formula: $Var(X) = E(X^2) - [E(X)]^2$. Substituting our values:

$$
Var(X) = \frac{11}{15} - \left(\frac{3}{5}ight)^2 = \frac{11}{15} - \frac{9}{25}
$$

To subtract these, we find the common denominator, which is $75$. The expression becomes:

$$
\frac{11 	imes 5 - 9 	imes 3}{75} = \frac{55 - 27}{75} = \frac{28}{75}
$$

And there it is. The variance is **$\frac{28}{75}$**. You have successfully navigated the logic of probability. Remember, math is not about memorizing formulas; it is about building a mental model of reality.

Animated Solution for Mathematics - Probability: A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let $X$ denote the number of defective pens. Then the variance of $X$ is

Select Answer:

Visualized Solution

Define the Random Variable $X$

Calculate Total Sample Space $n (S)$

Probability for $X = 0$

Probability for $X = 1$

Probability for $X = 2$

Formula for Expectation $E (X)$

Calculate Expectation $E (X)$

Formula for $E (X^{2})$

Calculate $E (X^{2})$

Formula for Variance $V a r (X)$

Substitute Values into Variance

Final Calculation

The Sigma Insight: Random Variables and Probability Distributions

The Geometry of Uncertainty

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

Define the Random Variable .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

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

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

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

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

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

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

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

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

Formula for Variance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Var(X)