Why can we write directly?

By definition, the variance is . Rearranging immediately gives , avoiding the separate subtraction and re-addition of .

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Statistics and Probability: Mathematics | JEE Main

Three 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\left(\mu^{2}+\sigma^{2}\right)

is equal to

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A250B25C30D20

Step-by-step solutionView

Correct answer

Using the relation $\mu^2 + \sigma^2 = E(X^2)$ , the expected value $E(X^2)$ evaluates to $2$ , which yields $10(\mu^2 + \sigma^2) = 20$ .

Option analysis

Why each option works or fails

A · 250

Treating the draws as independent with replacement and miscalculating the scale factor or introducing an extra power of ten. Recognize that sampling is done without replacement, and directly simplify $\mu^2 + \sigma^2$ to $E(X^2)$ to prevent cascading arithmetic errors.

B · 25

Incorrectly evaluating $E(X^2)$ or confusing $\mu^2 + \sigma^2$ with $(\mu + \sigma)^2$ , leading to an arithmetic result of $2.5$ before multiplying by $10$ . Recall that $\sigma^2 = E(X^2) - \mu^2$ , which implies $\mu^2 + \sigma^2 = E(X^2)$ . Calculate $E(X^2)$ directly using the probability distribution or indicators.

C · 30

Miscalculating the distribution probabilities $P(X=r)$ by miscounting combinations such as $\binom{7}{4-r}\binom{3}{r}$ , yielding $E(X^2) = 3$ . Ensure the total sample space $\binom{10}{4} = 210$ is used correctly alongside the valid values $X \in \{0, 1, 2, 3\}$ .

D · 20

Correct option. Identifying that $\mu^2 + \sigma^2 = E(X^2)$ and accurately computing the second moment gives $E(X^2) = 2$ , so $10 \times 2 = 20$ . None. The solution accurately links variance, expectation, and hypergeometric properties.

Reviewed route

Solution

StepWorking

01given

Total apples $N = 10$ , consisting of $3$ rotten and $7$ good apples. $n = 4$ apples are drawn at random without replacement. $X$ is the random variable representing the number of rotten apples drawn. $\mu = E[X]$ and $\sigma^2 = \text{Var}(X)$ .

02goal

Find the value of $10(\mu^2 + \sigma^2) = 10 E[X^2]$ .

03approach

Since $\sigma^2 = E[X^2] - \mu^2$ , we have $\mu^2 + \sigma^2 = E[X^2]$ . Thus, we only need to compute $E[X^2] = \sum_{x=0}^3 x^2 P(X=x)$ . The probability distribution of $X$ follows a hypergeometric distribution: $P(X=x) = \frac{\binom{3}{x}\binom{7}{4-x}}{\binom{10}{4}}$ for $x \in \{0, 1, 2, 3\}$ .

04execute

Total possible outcomes: $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{24} = 210$ . Compute probabilities for non-zero values: $P(X=1) = \frac{\binom{3}{1}\binom{7}{3}}{210} = \frac{3 \times 35}{210} = \frac{105}{210} = \frac{1}{2}$ , $P(X=2) = \frac{\binom{3}{2}\binom{7}{2}}{210} = \frac{3 \times 21}{210} = \frac{63}{210} = \frac{3}{10}$ , $P(X=3) = \frac{\binom{3}{3}\binom{7}{1}}{210} = \frac{1 \times 7}{210} = \frac{7}{210} = \frac{1}{30}$ . Now evaluate $E[X^2] = 0^2 P(0) + 1^2 P(1) + 2^2 P(2) + 3^2 P(3)$ : $E[X^2] = 1 \cdot \frac{1}{2} + 4 \cdot \frac{3}{10} + 9 \cdot \frac{1}{30} = \frac{1}{2} + \frac{6}{5} + \frac{3}{10} = \frac{5 + 12 + 3}{10} = \frac{20}{10} = 2$ . Then, $10(\mu^2 + \sigma^2) = 10 E[X^2] = 10 \times 2 = 20$ .

✓verify

Check via mean and variance separately: $\mu = 0 + 1(1/2) + 2(3/10) + 3(1/30) = 1/2 + 3/5 + 1/10 = 6/5$ . Then $\sigma^2 = E[X^2] - \mu^2 = 2 - 36/25 = 14/25$ . Thus, $10(\mu^2 + \sigma^2) = 10(36/25 + 14/25) = 10(50/25) = 20$ . Consistent.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can we write

\mu^2 + \sigma^2 = E[X^2]

directly?

By definition, the variance is $\sigma^2 = E[X^2] - (E[X])^2 = E[X^2] - \mu^2$ . Rearranging immediately gives $\mu^2 + \sigma^2 = E[X^2]$ , avoiding the separate subtraction and re-addition of $\mu^2$ .

P(I_2 = 1)

really

3/10

even without replacement?

Yes. By symmetry or the law of total probability, every position in a draw without replacement has the same marginal probability. For each draw, this probability of being rotten is $3/10$ .

Answer

Using the relation $\mu^2 + \sigma^2 = E(X^2)$ , the expected value $E(X^2)$ evaluates to $2$ , which yields $10(\mu^2 + \sigma^2) = 20$ .

Why each option works or fails

A: 250 - Treating the draws as independent with replacement and miscalculating the scale factor or introducing an extra power of ten. Recognize that sampling is done without replacement, and directly simplify $\mu^2 + \sigma^2$ to $E(X^2)$ to prevent cascading arithmetic errors.
B: 25 - Incorrectly evaluating $E(X^2)$ or confusing $\mu^2 + \sigma^2$ with $(\mu + \sigma)^2$ , leading to an arithmetic result of $2.5$ before multiplying by $10$ . Recall that $\sigma^2 = E(X^2) - \mu^2$ , which implies $\mu^2 + \sigma^2 = E(X^2)$ . Calculate $E(X^2)$ directly using the probability distribution or indicators.
C: 30 - Miscalculating the distribution probabilities $P(X=r)$ by miscounting combinations such as $\binom{7}{4-r}\binom{3}{r}$ , yielding $E(X^2) = 3$ . Ensure the total sample space $\binom{10}{4} = 210$ is used correctly alongside the valid values $X \in \{0, 1, 2, 3\}$ .
D · correct: 20 - Correct option. Identifying that $\mu^2 + \sigma^2 = E(X^2)$ and accurately computing the second moment gives $E(X^2) = 2$ , so $10 \times 2 = 20$ . None. The solution accurately links variance, expectation, and hypergeometric properties.

Step-by-step solution

given: Total apples $N = 10$ , consisting of $3$ rotten and $7$ good apples. $n = 4$ apples are drawn at random without replacement. $X$ is the random variable representing the number of rotten apples drawn. $\mu = E[X]$ and $\sigma^2 = \text{Var}(X)$ .
goal: Find the value of $10(\mu^2 + \sigma^2) = 10 E[X^2]$ .
approach: Since $\sigma^2 = E[X^2] - \mu^2$ , we have $\mu^2 + \sigma^2 = E[X^2]$ . Thus, we only need to compute $E[X^2] = \sum_{x=0}^3 x^2 P(X=x)$ . The probability distribution of $X$ follows a hypergeometric distribution: $P(X=x) = \frac{\binom{3}{x}\binom{7}{4-x}}{\binom{10}{4}}$ for $x \in \{0, 1, 2, 3\}$ .
execute: Total possible outcomes: $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{24} = 210$ . Compute probabilities for non-zero values: $X$ , $P(X=2) = \frac{\binom{3}{2}\binom{7}{2}}{210} = \frac{3 \times 21}{210} = \frac{63}{210} = \frac{3}{10}$ , $P(X=3) = \frac{\binom{3}{3}\binom{7}{1}}{210} = \frac{1 \times 7}{210} = \frac{7}{210} = \frac{1}{30}$ . Now evaluate $E[X^2] = 0^2 P(0) + 1^2 P(1) + 2^2 P(2) + 3^2 P(3)$ : $E[X^2] = 1 \cdot \frac{1}{2} + 4 \cdot \frac{3}{10} + 9 \cdot \frac{1}{30} = \frac{1}{2} + \frac{6}{5} + \frac{3}{10} = \frac{5 + 12 + 3}{10} = \frac{20}{10} = 2$ . Then, $10(\mu^2 + \sigma^2) = 10 E[X^2] = 10 \times 2 = 20$ .
verify: Check via mean and variance separately: $\mu = 0 + 1(1/2) + 2(3/10) + 3(1/30) = 1/2 + 3/5 + 1/10 = 6/5$ . Then $\sigma^2 = E[X^2] - \mu^2 = 2 - 36/25 = 14/25$ . Thus, $10(\mu^2 + \sigma^2) = 10(36/25 + 14/25) = 10(50/25) = 20$ . Consistent.

Shortcut:

When to use it: Fastest conceptual method: write $X$ as a sum of Bernoulli indicators without evaluating the full distribution table.

given: A sample of size $4$ is drawn without replacement from $3$ rotten and $7$ good apples. $X$ is the number of rotten apples.

approach: Note that $\mu^2 + \sigma^2 = E[X^2]$ . Let $X = I_1 + I_2 + I_3 + I_4$ , where $I_k = 1$ if the $k$ -th drawn apple is rotten, and $0$ otherwise. By symmetry, $P(I_k = 1) = \frac{3}{10}$ and $P(I_j = 1, I_k = 1) = \frac{3}{10} \times \frac{2}{9} = \frac{1}{15}$ for $j \neq k$ .

execute: We have $E[X^2] = E\left[\sum_{k=1}^4 I_k^2 + 2\sum_{1 \le j < k \le 4} I_j I_k\right]$ . Since $I_k^2 = I_k$ , this is: $3/10$ . Therefore, $10(\mu^2 + \sigma^2) = 10(2) = 20$ .

verify: Check that $E[X] = 4 \times (3/10) = 1.2$ , so $\mu^2 = 1.44$ . $E[X^2] = 2$ , so $\sigma^2 = 2 - 1.44 = 0.56 = 14/25$ . Both values match the exact distribution.

Statistics and Probability: Mathematics | JEE Main

What feels right?

Using the relation $\mu^2 + \sigma^2 = E(X^2)$ , the expected value $E(X^2)$ evaluates to $2$ , which yields $10(\mu^2 + \sigma^2) = 20$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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