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JEE Mathematics 2023 Question with Solution

Three rotten apples are accidentally mixed with seven good apples, and four apples are drawn one by one without replacement. Let the random variable XX denote the number of rotten apples. If μ\mu and σ2\sigma^2 represent the mean and variance of XX, respectively, then 10(μ2+σ2)10(\mu^2 + \sigma^2) is equal to:

  • A

    2020

  • B

    250250

  • C

    2525

  • D

    3030

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: There are 33 rotten apples and 77 good apples, so total apples N=10N = 10. Four apples are drawn without replacement, so n=4n = 4. The random variable XX denotes the number of rotten apples drawn.

Find: The value of 10(μ2+σ2)10(\mu^2 + \sigma^2).

Since the draws are made without replacement, XX follows a hypergeometric distribution. Its probability mass function is

P(X=k)=(3k)(74−k)(104)P(X = k) = \frac{\binom{3}{k}\binom{7}{4-k}}{\binom{10}{4}}

For a hypergeometric distribution,

μ=nKN,σ2=nK(N−K)(N−n)N2(N−1)\mu = \frac{nK}{N}, \qquad \sigma^2 = \frac{nK(N-K)(N-n)}{N^2(N-1)}

where K=3K = 3, n=4n = 4, and N=10N = 10.

Now compute the mean:

μ=4×310=1210=1.2\mu = \frac{4 \times 3}{10} = \frac{12}{10} = 1.2

Now compute the variance:

σ2=4×3×(10−3)×(10−4)102×(10−1)=4×3×7×6100×9=504900=0.56\sigma^2 = \frac{4 \times 3 \times (10-3) \times (10-4)}{10^2 \times (10-1)} = \frac{4 \times 3 \times 7 \times 6}{100 \times 9} = \frac{504}{900} = 0.56

Therefore,

μ2=(1.2)2=1.44,μ2+σ2=1.44+0.56=2\mu^2 = (1.2)^2 = 1.44, \qquad \mu^2 + \sigma^2 = 1.44 + 0.56 = 2

So,

10(μ2+σ2)=10×2=2010(\mu^2 + \sigma^2) = 10 \times 2 = 20

Therefore, the value is 2020. The solution working gives 2020, which matches option A. The solution text stating option D / option (4) is inconsistent with the actual calculation.

Using Hypergeometric Mean and Variance

Given: Sampling is without replacement from 1010 apples containing 33 rotten ones.

Find: 10(μ2+σ2)10(\mu^2 + \sigma^2) for the number of rotten apples drawn in 44 draws.

Identify the parameters of the hypergeometric distribution:

  • Total population size: N=10N = 10
  • Number of rotten apples: K=3K = 3
  • Number of draws: n=4n = 4

Use the standard formulas:

μ=nKN=4⋅310=65\mu = \frac{nK}{N} = \frac{4 \cdot 3}{10} = \frac{6}{5} σ2=nK(N−K)(N−n)N2(N−1)=4⋅3⋅7⋅6102⋅9=504900=1425\sigma^2 = \frac{nK(N-K)(N-n)}{N^2(N-1)} = \frac{4 \cdot 3 \cdot 7 \cdot 6}{10^2 \cdot 9} = \frac{504}{900} = \frac{14}{25}

Now square the mean:

μ2=(65)2=3625\mu^2 = \left(\frac{6}{5}\right)^2 = \frac{36}{25}

Then add:

μ2+σ2=3625+1425=5025=2\mu^2 + \sigma^2 = \frac{36}{25} + \frac{14}{25} = \frac{50}{25} = 2

Finally,

10(μ2+σ2)=10⋅2=2010(\mu^2 + \sigma^2) = 10 \cdot 2 = 20

Hence, the correct option is A.

Common mistakes

  • Treating the distribution as binomial instead of hypergeometric. That is wrong because the apples are drawn without replacement. Use hypergeometric formulas for mean and variance.

  • Using the mean formula correctly but forgetting the finite population correction in the variance. For sampling without replacement, the variance is not the binomial variance npqnpq; use the hypergeometric variance formula.

  • Trusting the mislabeled option statement on the solution instead of the actual calculation. The worked value comes out to 2020, so the defensible option is the one containing 2020.

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