Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not

Q: Why can't we just use the Binomial Distribution formula $Var(X) = npq$?

That is a fantastic question! The Binomial Distribution applies only when trials are independent, which happens in sampling *with* replacement. Here, we are sampling *without* replacement. The probability of picking a defective orange changes after the first draw, so we must use the Hypergeometric approach (calculating probabilities directly) instead of the Binomial shortcut.

Q: Why do we subtract the square of the mean at the end?

Think of the variance as a measure of 'spread' or 'uncertainty.' The formula $Var(X) = E(X^2) - [E(X)]^2$ is the mathematical definition of this spread. We calculate the average of the squares ($E(X^2)$) and then remove the 'bias' of the mean squared to isolate the pure variance. It is the standard, most robust way to calculate variance in discrete probability.

Q: Is there a faster way to find $E(X^2)$ without doing the full summation?

For small values like $X \in \{0, 1, 2\}$, the direct summation is actually the fastest and safest method. While there are advanced formulas for the variance of a hypergeometric distribution, deriving them often takes longer than simply computing the three terms. Stick to the fundamental definition; it prevents silly algebraic errors.

Question

Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is

Options:

(A) $28/75$

(B) $18/25$

(C) $26/75$

(D) $14/25$

Sigma Percentile · Accepted Answer

### The Basket of Uncertainty: A Journey into Probability

Welcome, future engineer. Today, we are not just solving a probability problem; we are peeling back the layers of uncertainty.

Imagine you are standing in a market, holding a basket of ten oranges. Seven are perfect, and three are defective. You cannot tell them apart.

You reach in and pull out two. This is the essence of life—making decisions with incomplete information. To master this, we must quantify that uncertainty using the language of statistics.

### Phase 1: Defining the Random Variable

Before we touch the math, we must define our universe. We have a total of $10$ oranges, and we are drawing $n = 2$.

Let $X$ be our random variable, representing the number of defective oranges we hold in our hand. Since we only draw two, $X$ can only take three possible values: $0$ (both are good), $1$ (one good, one defective), or $2$ (both are defective).

This is our sample space. It is finite, discrete, and perfectly manageable.

### Phase 2: The Probability Distribution

To calculate the variance, we first need the probability of each outcome. We use combinations because the order of selection does not matter.

The total number of ways to choose $2$ oranges from $10$ is given by:

$$^{10}C_2 = \frac{10 	imes 9}{2 	imes 1} = 45$$

This $45$ is our denominator for every probability.

For $X=0$, we need to choose $2$ good oranges from the $7$ available:

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

For $X=1$, we need $1$ defective (from $3$) and $1$ good (from $7$):

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

For $X=2$, we need $2$ defective (from $3$):

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

Always perform the 'sanity check': $\frac{7}{15} + \frac{7}{15} + \frac{1}{15} = \frac{15}{15} = 1$. The universe is in balance.

### Phase 3: The Mean and the Second Moment

Now, we calculate the expected value, $E(X)$, which is the weighted average of our outcomes:

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

Next, we need the second moment, $E(X^2)$, to measure the 'spread.' We square the values of $X$ before multiplying by the probabilities:

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

### Phase 4: The Final Calculation

We have arrived at the final gate. The variance is defined as $Var(X) = E(X^2) - [E(X)]^2$. This formula is the bridge between the average and the spread.

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 $75$:

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

There it is. The **variance is $\frac{28}{75}$**.

You have successfully navigated the probability space, accounted for the dependencies, and arrived at the precise measure of dispersion. Remember, in JEE Advanced, it is not just about the answer; it is about the clarity of your logic. Keep this rigor in your toolkit, and no problem will ever be too complex.

Select Answer:

Visualized Solution

Visualizing the Setup

Defining the Random Variable $X$

Total Possible Outcomes

Calculating $P (X = 0)$

Calculating $P (X = 1)$

Calculating $P (X = 2)$

The Probability Table

Logic Bridge: Formula for Mean $E (X)$

Computing Mean $E (X)$

Logic Bridge: Formula for $E (X^{2})$

Computing $E (X^{2})$

Logic Bridge: Variance Formula

Final Computation of Variance

The Way Forward

The Sigma Insight: Random Variables and Probability Distributions

The Basket of Uncertainty

Phase 1

Phase 2

Phase 3

Phase 4

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