Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable x to

Q: Why can't I use the Binomial Distribution formula here? Isn't it just picking apples?

That is a classic trap! The Binomial Distribution requires independent trials with a constant probability of success (sampling with replacement). Here, the probability of picking a rotten apple changes after the first draw because we are sampling without replacement. This is a Hypergeometric distribution scenario, so we must use combinations.

Q: Why is the variance formula $E(X^2) - [E(X)]^2$ better than the definition?

It is computationally superior. Calculating the mean $\mu$ first and then subtracting it from every $x_i$ before squaring often leads to messy fractions and sign errors. The $E(X^2) - [E(X)]^2$ method keeps the arithmetic cleaner until the very last step, minimizing the chance of a silly mistake.

Q: Should I simplify the fractions at every step? I feel like I'm wasting time.

Absolutely not! Keep the denominator as $153$ throughout your calculations. Simplifying early often leads to different denominators, making the final subtraction of $E(X^2) - [E(X)]^2$ a nightmare. Keep the common denominator until the very end; it is the secret to speed and accuracy.

Question

Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable $x$ to be the number of rotten apples in a draw of two apples, the variance of $x$ is

Options:

(A) $\frac{37}{153}$

(B) $\frac{57}{153}$

(C) $\frac{47}{153}$

(D) $\frac{40}{153}$

Sigma Percentile · Accepted Answer

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

Imagine you are standing in a quiet room, a basket of eighteen apples before you. Fifteen are crisp, red, and perfect. Three are rotten, hidden amongst the good ones.

You are tasked with drawing two apples. In the high-stakes environment of the JEE Advanced, this isn't just a basket of fruit; it is a microcosm of probability theory. We are dealing with a discrete random variable, $X$, representing the number of rotten apples in our draw.

Let us embark on this journey to find the variance of $X$, a measure of how much our results might deviate from the average.

### Phase 1: The Foundation of Combinations

Before we can predict the future, we must understand the total possibilities. We are drawing two apples from eighteen. This is a classic combination problem.

We do not care about the order in which we pick the apples, only the final composition of our pair. Thus, we use the combination formula $\binom{n}{r}$.

The total number of ways to choose two apples from eighteen is:

$$
\binom{18}{2} = \frac{18 	imes 17}{2 	imes 1} = 153
$$

This number, $153$, is our anchor. It will be the denominator for every probability we calculate. Do not simplify it yet; keep it as your constant companion.

### Phase 2: Mapping the Probability Distribution

Now, let us define our random variable $X$. Since we are drawing two apples, $X$ can only take the values $0, 1,$ or $2$.

For $X=0$ (zero rotten apples), we must pick two good apples from the fifteen available. The number of ways is:

$$
P(X=0) = \frac{\binom{3}{0} 	imes \binom{15}{2}}{153} = \frac{1 	imes 105}{153} = \frac{105}{153}
$$

For $X=1$ (one rotten, one good), we pick one from the three rotten and one from the fifteen good. The number of ways is:

$$
P(X=1) = \frac{\binom{3}{1} 	imes \binom{15}{1}}{153} = \frac{3 	imes 15}{153} = \frac{45}{153}
$$

For $X=2$ (two rotten apples), we pick two from the three rotten. The number of ways is:

$$
P(X=2) = \frac{\binom{3}{2} 	imes \binom{15}{0}}{153} = \frac{3 	imes 1}{153} = \frac{3}{153}
$$

Notice the elegance here: $105 + 45 + 3 = 153$. The sum of probabilities is exactly $1$. The universe of our problem is balanced.

### Phase 3: The Expectation and the Variance

To find the variance, we need two pillars: the expected value $E(X)$ and the expected value of the square $E(X^2)$.

First, the mean, $E(X) = \sum x_i P(x_i)$. Calculating this:

$$
E(X) = 0\left(\frac{105}{153}ight) + 1\left(\frac{45}{153}ight) + 2\left(\frac{3}{153}ight) = \frac{45 + 6}{153} = \frac{51}{153} = \frac{1}{3}
$$

Next, the second moment, $E(X^2) = \sum x_i^2 P(x_i)$. Calculating this:

$$
E(X^2) = 0^2\left(\frac{105}{153}ight) + 1^2\left(\frac{45}{153}ight) + 2^2\left(\frac{3}{153}ight) = \frac{45 + 12}{153} = \frac{57}{153}
$$

### Phase 4: The Final Synthesis

We arrive at the final step, the variance formula: $	ext{Var}(X) = E(X^2) - [E(X)]^2$. Substituting our values, we get:

$$
	ext{Var}(X) = \frac{57}{153} - \left(\frac{1}{3}ight)^2 = \frac{57}{153} - \frac{1}{9}
$$

To subtract these, we convert $\frac{1}{9}$ to $\frac{17}{153}$. Finally:

$$
	ext{Var}(X) = \frac{57 - 17}{153} = \mathbf{\frac{40}{153}}
$$

There it is. The chaos of the rotten apples is tamed by the rigor of mathematics. You have successfully navigated the distribution, the expectation, and the variance.

Remember this process; it is the same logic that governs everything from quantum mechanics to financial risk. You are not just solving a problem; you are mastering the language of uncertainty.

Animated Solution for Mathematics - Probability: Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable $x$ to be the number of rotten apples in a draw of two apples, the variance of $x$ is

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Visualized Solution

Problem Setup: Counting the Apples

Defining the Random Variable $X$

Calculating Total Outcomes $(2 18)$

Probability of Zero Rotten Apples $P (X = 0)$

Probability of One Rotten Apple $P (X = 1)$

Probability of Two Rotten Apples $P (X = 2)$

Calculating the Expectation $E (X)$

Calculating $E (X^{2})$

The Variance Formula

Final Calculation: Subtracting Fractions

The Sigma Insight: Random Variables and Probability Distributions

The Basket of Uncertainty

Phase 1

Phase 2

Phase 3

Phase 4

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