There rotten apples are mixed accidently with seven good apples and four apples are drawn one by one

Q: Wait, why can't we just 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. Because we are drawing apples *without* replacement, the probability of picking a rotten apple changes with every draw. This dependency makes it a Hypergeometric distribution, not a binomial one.

Q: Why does $\mu^2 + \sigma^2$ equal $E(X^2)$? Is that always true?

Yes, it is always true! By definition, the variance $\sigma^2$ is the expected value of the squared deviation from the mean: $\sigma^2 = E[(X - \mu)^2]$. If you expand that expression, you get $E[X^2 - 2X\mu + \mu^2]$, which simplifies to $E(X^2) - 2\mu E(X) + \mu^2$. Since $E(X) = \mu$, this becomes $E(X^2) - 2\mu^2 + \mu^2$, which is $E(X^2) - \mu^2$. Rearranging gives $\mu^2 + \sigma^2 = E(X^2)$.

Q: Do I really need to calculate all the probabilities individually?

For this specific problem, yes, it is the most reliable path. Since the random variable $X$ only takes values $0, 1, 2,$ and $3$, calculating the four probabilities is quite fast. Trying to use the general formula for the variance of a Hypergeometric distribution is possible, but it is often more prone to algebraic errors than simply summing the moments directly.

Question

There rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable $X$ denote the number of rotten apples. If $\mu$ and $\sigma^2$ represent mean and variance of $X$, respectively, then $10(\mu^2 + \sigma^2)$ is equal to

Options:

(A) 20

(B) 250

(C) 25

(D) 30

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a basket containing ten apples. Seven are fresh, and three are rotten. We are drawing $4$ apples one by one without replacement, which is a classic scenario following the **Hypergeometric distribution**.

Our goal is to find the value of $10(\mu^2 + \sigma^2)$, where $\mu$ is the mean and $\sigma^2$ is the variance of the number of rotten apples, $X$.

The total number of ways to choose $4$ apples out of $10$ is given by the combination formula:

$$ \binom{10}{4} = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210 $$

This value, $210$, serves as the denominator for all our probability calculations.

### Mapping the Distribution

The random variable $X$ represents the number of rotten apples drawn. Since there are only $3$ rotten apples available, $X$ can take the values $0, 1, 2,$ or $3$.

We calculate the probability for each case as follows:

For $X=0$:
$$ P(X=0) = \frac{\binom{3}{0} 	imes \binom{7}{4}}{210} = \frac{1 	imes 35}{210} = \frac{1}{6} $$

For $X=1$:
$$ P(X=1) = \frac{\binom{3}{1} 	imes \binom{7}{3}}{210} = \frac{3 	imes 35}{210} = \frac{1}{2} $$

For $X=2$:
$$ P(X=2) = \frac{\binom{3}{2} 	imes \binom{7}{2}}{210} = \frac{3 	imes 21}{210} = \frac{3}{10} $$

For $X=3$:
$$ P(X=3) = \frac{\binom{3}{3} 	imes \binom{7}{1}}{210} = \frac{1 	imes 7}{210} = \frac{1}{30} $$

### The Algebraic Shortcut

The question asks for $10(\mu^2 + \sigma^2)$. Instead of calculating the mean $\mu = E(X)$ and the variance $\sigma^2 = E(X^2) - \mu^2$ separately, we can simplify the expression.

Substituting the definition of variance into the target expression:

$$ \mu^2 + \sigma^2 = \mu^2 + (E(X^2) - \mu^2) = E(X^2) $$

The $\mu^2$ terms cancel out perfectly. We only need to calculate the second moment, $E(X^2)$.

### Final Calculation

We compute $E(X^2)$ using the formula $E(X^2) = \sum x^2 P(X=x)$:

$$ E(X^2) = 0^2\left(\frac{1}{6}ight) + 1^2\left(\frac{1}{2}ight) + 2^2\left(\frac{3}{10}ight) + 3^2\left(\frac{1}{30}ight) $$

Simplifying the terms:

$$ E(X^2) = 0 + \frac{1}{2} + \frac{12}{10} + \frac{9}{30} $$

Using a common denominator of $30$:

$$ E(X^2) = \frac{15 + 36 + 9}{30} = \frac{60}{30} = 2 $$

Finally, we multiply by $10$ as requested:

$$ 10 	imes E(X^2) = 10 	imes 2 = \mathbf{20} $$

Select Answer:

Visualized Solution

Problem Setup & Random Variable $X$

Total Sample Space

Probability for $X = 0$

Probability for $X = 1$

Probability for $X = 2$

Probability for $X = 3$

The Variance Shortcut

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

Calculating $E (X^{2})$

Final Computation

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

Mapping the Distribution

The Algebraic Shortcut

Final Calculation

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