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Indian Navy 10+2 B.Tech Cadet Statistics and Probability — practice questions

Explore 30 free Statistics and Probability questions in Indian Navy 10+2 B.Tech Cadet Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Statistics and Probability questions and exercises with answers

Showing questions 1–30 of 30.

Mutually exclusive events A and B:Read the answer and worked solution →P(A) = 1/3, P(B) = 1/2, P(A ∩ B) = 1/6. P(A ∪ B) =Read the answer and worked solution →For two events A and B, P(A | B) = 0.6, P(B) = 0.5, P(A) = 0.4. P(B | A) =Read the answer and worked solution →Chebyshev's inequality: P(|X - μ| ≥ kσ) ≤Read the answer and worked solution →Tap every measure of dispersion your syllabus listsRead the answer and worked solution →Central tendency or dispersion?Read the answer and worked solution →Complete the definition of rangeRead the answer and worked solution →Working out a rangeRead the answer and worked solution →Order the argument: why mean deviation uses absolute valuesRead the answer and worked solution →True or false: why plain deviations failRead the answer and worked solution →Complete the definition of mean deviationRead the answer and worked solution →Match each measure to how it is builtRead the answer and worked solution →True or false: variance and standard deviationRead the answer and worked solution →Squared or absolute?Read the answer and worked solution →Reading the definitions backRead the answer and worked solution →True or false: what this chapter coversRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Order the steps: compute a mean deviation about the meanRead the answer and worked solution →Order the steps: compute a standard deviationRead the answer and worked solution →Flip and pair: symbol with meaningRead the answer and worked solution →Grouped data or ungrouped data?Read the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Tap every statement true of the rangeRead the answer and worked solution →Complete the reason absolute values appearRead the answer and worked solution →Central value choicesRead the answer and worked solution →Assertion and reason: choosing a measureRead the answer and worked solution →True or false: measures comparedRead the answer and worked solution →Match each idea to what it answersRead the answer and worked solution →

Worked practice questions

Mutually exclusive events A and B:

  1. A = B
  2. P(A)P(B)
  3. Independent
  4. P(A ∩ B) = 0; P(A ∪ B) = P(A) + P(B)
Show answer and solution

Correct answer: D. P(A ∩ B) = 0; P(A ∪ B) = P(A) + P(B)

**Mutually exclusive:** cannot occur together. P(A ∩ B) = 0. P(A ∪ B) = P(A) + P(B). NOT same as independent (mutually exclusive non-zero events are dependent).

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P(A) = 1/3, P(B) = 1/2, P(A ∩ B) = 1/6. P(A ∪ B) =

  1. 1
  2. 1/3 + 1/2 - 1/6 = 2/3
  3. 1/2
  4. 5/6
Show answer and solution

Correct answer: B. 1/3 + 1/2 - 1/6 = 2/3

Inclusion-exclusion: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 2/6 + 3/6 - 1/6 = 4/6 = 2/3.

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For two events A and B, P(A | B) = 0.6, P(B) = 0.5, P(A) = 0.4. P(B | A) =

  1. 0.75
  2. 0.4
  3. 0.5
  4. 0.6 × 0.5 / 0.4 = 0.75 (Bayes)
Show answer and solution

Correct answer: D. 0.6 × 0.5 / 0.4 = 0.75 (Bayes)

We have P(B|A) = P(A|B) × P(B)/P(A). That is P(B|A) = 0.6 × 0.5 / 0.4. So P(B|A) = 0.75.

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