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BIHAR CET Vectors & 3D Geometry — practice questions — page 4

Explore 108 free Vectors & 3D Geometry questions in BIHAR CET Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Vectors & 3D Geometry questions and exercises with answers

Showing questions 91–108 of 108.

Assertion and reasonRead the answer and worked solution →Tap every statement that follows from independenceRead the answer and worked solution →The multiplication ruleRead the answer and worked solution →Spot the one that does not belongRead 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 →Complete the theorem of total probabilityRead the answer and worked solution →Which theorem answers which question?Read the answer and worked solution →True or false: total probability and BayesRead the answer and worked solution →Match each historical figure to their contributionRead the answer and worked solution →The Historical NoteRead the answer and worked solution →Flip and pair: notation with meaningRead the answer and worked solution →Order the reasoning: from a partition to a posteriorRead the answer and worked solution →Tap every quantity that is a probabilityRead the answer and worked solution →Spot the odd one outRead the answer and worked solution →Assertion and reason: the multiplication ruleRead the answer and worked solution →True or false: the Historical NoteRead the answer and worked solution →Sort each statement by which idea it describesRead the answer and worked solution →

Worked practice questions

Assertion and reason

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Questions
  1. Reason
    P(F) sits in the denominator of the definition.
    Verdict
    Both are true; the reason explains the assertion.
    Assertion
    P(E|F) requires P(F) ≠ 0.
  2. Reason
    The denominator sums over every branch, so the branches must cover S once.
    Verdict
    Both are true; the reason explains the assertion.
    Assertion
    Bayes' theorem needs the events to form a partition.
  3. Reason
    Mutually exclusive events satisfy P(E ∩ F) = 0.
    Verdict
    The assertion is false and the reason is true.
    Assertion
    Two mutually exclusive events with nonzero probability are independent.
  4. Reason
    Conditioning on F still leaves a complete probability model.
    Verdict
    Both are true; the reason explains the assertion.
    Assertion
    P(E′|F) = 1 − P(E|F).
The reason must actually explain the assertion. Source: syllabus Maths Part II, Ch 13 Probability.

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Tap every statement that follows from independence

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Questions
  1. Items
    1. Text
      P(E ∩ F) = P(E) P(F)
      Matches the category
      Yes
    2. Text
      P(E|F) = P(E)
      Matches the category
      Yes
    3. Text
      P(F|E) = P(F)
      Matches the category
      Yes
    4. Text
      P(E ∩ F) = 0
      Matches the category
      No
    5. Text
      P(E) + P(F) = 1
      Matches the category
      No
    Category
    Follows from E and F being independent
  2. Items
    1. Text
      The events partition the sample space
      Matches the category
      Yes
    2. Text
      They are pairwise disjoint
      Matches the category
      Yes
    3. Text
      Each has nonzero probability
      Matches the category
      Yes
    4. Text
      They are all equally likely
      Matches the category
      No
    5. Text
      They are independent of each other
      Matches the category
      No
    Category
    Needed for Bayes' theorem to apply
Three equivalent statements of independence, set against results that need something else. Source: syllabus Maths Part II, Ch 13 Probability.

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The multiplication rule

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Questions
  1. Question
    P(E ∩ F) equals
    Choices
    1. P(E) + P(F|E)
    2. P(E) ÷ P(F)
    3. P(E) P(F|E)
    Correct choice
    P(E) P(F|E)
  2. Question
    P(E ∩ F) also equals
    Choices
    1. P(F) P(E|F)
    2. P(F) − P(E)
    3. P(E|F)
    Correct choice
    P(F) P(E|F)
  3. Question
    The multiplication rule needs
    Choices
    1. the events to be independent
    2. the conditioning event to have nonzero probability
    3. a partition
    Correct choice
    the conditioning event to have nonzero probability
Two forms of the same rule, depending on which event you condition on. Source: syllabus Maths Part II, Ch 13 Probability.

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