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Probability and Statistics

GATE CSE 2024 Set 1 · Q27

MSQ1 markEasy

Time

Let AA and BB be two events in a probability space with P(A)=0.3P(A) = 0.3, P(B)=0.5P(B) = 0.5, and P(A∩B)=0.1P(A \cap B) = 0.1. Which of the following statements is/are TRUE?

Select all that applyOne or more correct

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Solution

Step-by-step verified solution

Official answer: B) P(A ∪ B) = 0.7; C) P(A ∩ Bc) = 0.2, where Bc is the complement of the event B

Probability and Statistics

We are given:

  • P(A)=0.3P(A) = 0.3

  • P(B)=0.5P(B) = 0.5

  • P(A∩B)=0.1P(A \cap B) = 0.1

Let us evaluate each statement:

  1. Option A: For two events to be independent, P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B). Here, P(A)×P(B)=0.3×0.5=0.15≠0.1P(A) \times P(B) = 0.3 \times 0.5 = 0.15 \neq 0.1. Hence, AA and BB are not independent. *(Statement A is false)*

  2. Option B: By the principle of inclusion-exclusion:

    P(A∪B)=P(A)+P(B)−P(A∩B)=0.3+0.5−0.1=0.7P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.5 - 0.1 = 0.7

    *(Statement B is true)*

  3. Option C: Using the complement rule for sets:

    P(A∩Bc)=P(A)−P(A∩B)=0.3−0.1=0.2P(A \cap B^c) = P(A) - P(A \cap B) = 0.3 - 0.1 = 0.2

    *(Statement C is true)*

  4. Option D: By De Morgan's Law:

    P(Ac∩Bc)=P((A∪B)c)=1−P(A∪B)=1−0.7=0.3≠0.4P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B) = 1 - 0.7 = 0.3 \neq 0.4

    *(Statement D is false)*

Thus, the correct options are B and C.

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This Probability and Statistics previous year question is Q27 of the GATE CSE 2024 Set 1 paper. Find more GATE CSE PYQs on Probability and Statistics in Engineering Mathematics, or browse all GATE CSE PYQs.

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GateAI solution cardGATE CSE 2024 Set 1 · Q27 · MSQ

Let A and B be two events in a probability space with P(A) = 0.3, P(B) = 0.5, and P(A ∩ B) = 0.1. Which of the following statements is/are TRUE?

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