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If E and F are independent events such that 0 < P(E) <1 and 0 < P(F) < 1, then (1989 - 2 Marks)
  • a)
    E and F are mutually exclusive
  • b)
    E and Fc ( the complement of the event F) are independent
  • c)
    Ec and Fc are independent
  • d)
    P(E | F) + P(Ec | F) = 1.
Correct answer is option 'B,C,D'. Can you explain this answer?
Verified Answer
If E and F are independent events such that 0 < P(E) <1 and 0 &l...
Since E and F are independent
∴ P (E ∩ F) = P (E) . P ( F) ...(1)
Now, P (E ∩ Fc) = P (E) – P (E ∩ F) = P (E) – P(E) P(F) [Using (1)]
= P (E) [1 – P (F)] = P (E) P (Fc)
∴ E and Fc are independent.
Again P (Ec ∩ Fc) = P (E ∪ F)c = 1 – P (E ∪ F) = 1– P (E) – P (F) + P (E ∩ F)
= 1 – P (E) – P (F) + P (E) P (F)
= ((1– P (E) (1 – P (F)) = P (Ec) P (Fc)
∴ Ec and Fc are independent.
Also P (E/ F) + P (Ec/F)
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If E and F are independent events such that 0 < P(E) <1 and 0 < P(F) < 1, then (1989 - 2 Marks)a)E and F are mutually exclusiveb)E and Fc ( the complement of the event F) are independentc)Ec and Fc are independentd)P(E | F) + P(Ec | F) = 1.Correct answer is option 'B,C,D'. Can you explain this answer?
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If E and F are independent events such that 0 < P(E) <1 and 0 < P(F) < 1, then (1989 - 2 Marks)a)E and F are mutually exclusiveb)E and Fc ( the complement of the event F) are independentc)Ec and Fc are independentd)P(E | F) + P(Ec | F) = 1.Correct answer is option 'B,C,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If E and F are independent events such that 0 < P(E) <1 and 0 < P(F) < 1, then (1989 - 2 Marks)a)E and F are mutually exclusiveb)E and Fc ( the complement of the event F) are independentc)Ec and Fc are independentd)P(E | F) + P(Ec | F) = 1.Correct answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If E and F are independent events such that 0 < P(E) <1 and 0 < P(F) < 1, then (1989 - 2 Marks)a)E and F are mutually exclusiveb)E and Fc ( the complement of the event F) are independentc)Ec and Fc are independentd)P(E | F) + P(Ec | F) = 1.Correct answer is option 'B,C,D'. Can you explain this answer?.
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