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Mathematics · Probability

JEE Main 2025 — 28 January, Morning Shift — Question 19

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 xx is :

  1. Option A:

    28/7528 / 75

    Correct
  2. Option B:

    14/2514 / 25

  3. Option C:

    26/7526 / 75

  4. Option D:

    18/2518 / 25

Answer: A

Step-by-step solution

figure

Now, μ=∑xipi=4290+1290=5490\mu=\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{42}{90}+\frac{12}{90}=\frac{54}{90}

σ2=∑pix12−μ2=4290+2490−(5490)2\sigma^{2}=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{1}^{2}-\mu^{2}=\frac{42}{90}+\frac{24}{90}-\left(\frac{54}{90}\right)^{2}

⇒6690−(5490)2\Rightarrow \frac{66}{90}-\left(\frac{54}{90}\right)^{2}

σ2⇒2875\sigma^{2} \Rightarrow \frac{28}{75}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
28 January, Morning Shift
Subject
Mathematics
Chapter
Probability
Topic
Random Variables, Binomial & Poission Distribution

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