Set theoretic principles Assignment Help

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Set theoretic principles:

If 'A' and 'B' be any 2 events of sample space then

  • A∪B would stand for the occurrence of at least 1 of them.
  • A∩B stands for the simultaneous occurrence of A and B.
  • 556_Set theoretic principles.png  (or A') stands for the non occurrence of A
  • 2186_Set theoretic principles1.png (or A' ∩ B') stands for non occurrence of A and B both.
  • A ⊆ B stands for 'occurrence of A implies occurrence of B'.
  • If A and B are any 2 events, then P(A∪B) = P(A) + P(B) - P(A∩B)
  • If A and B are mutually exclusive events, then P(A∪B) = P(A) + P(B).

            In this case there will not be any sample point present in A∩B.

  • P(A') = 1 - P(A)

            As, A∪A' = S and A and A' are mutually exclusive events

            =>  P(A∪A') = P(A) + P(A') = P(S) = 1 =>  P(A') = 1 - P(A).

  • P(A ∩ B') = P(A) - P(A ∩ B)

            As A∩B' and A∩B are mutually exclusive events and (A∩B')∪(A∩B) = A

            => P(A∩B') + P(A∩B) = P(A) => P(A∩B') = P(A) - P(A∩B)

            Likewise, P(A'∩B) = P(B) - P(A∩B)

  • P(A'∪B') = 1 - P(A ∩ B)

            As,  (A'∪B') ∪ (A∩B) = S

            =>  P(A'∪B') + P(A ∩ B) = 1 =>  P(A' ∪ B') = 1 - P(A∩B)

            Likewise, P(A' ∩ B') = 1 - P(A∪B)

  • P(exactly one of A, B occurs)

            = P(A∩B') + P(A'∩B) = P(A) + P(B) - 2P(A∩B)

            = P(A∪B) - P(A∩B) = P(A' ∪ B') - P(A' ∩ B')

            If A, B, C are any 3 events of sample space, then

  • P(A∪B∪C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)
  • P (Exactly 1 of A, B, C occurs)

            = P(A) + P(B) + P(C) - 2P(A∩B) - 2A(A∩C) - 2P(B∩C) + 3P(A∩B∩C)

  • P(Exactly 2 of A, B, C occur) = P(A ∩ B) + P(B ∩ C) + P(A ∩ C) - 3P(A ∩ B ∩ C)
  • P (at least 2 of A, B, C occur) = P(A ∩ B) + P(B ∩ C) + P(A ∩ C) - 2P(A ∩ B ∩ C)
  • If A1, A2 L , An are ¢n' events, then P(A1∪ A2 L ∪An)

           1201_Set theoretic principles3.png

  • P (A ∪ B) ≥ max (P(A), P(B), P(A) + P(B) - 1)

            as A ⊆ A∪ B => P(A) ≤ P(A∪B)

            Like wise, B ⊆ A ∪ B => P(B) ≤ P(A∪B) =>  P(A∪B) ≥ Max (P(A), P(B))

            Also  P(A∪B) = P(A) + P(B) - P(A ∩ B)

            P(A) + P(B) - 1  ≤  P(A∪B) ≤ P(A) + P(B) (As 0 ≤ P(A ∩ B) ≤ 1)

            Max (P(A) + P(B) - 1,  P(A), P(B)) ≤ P(A∪B) ≤ P(A) + P(B)

  • If out of m + n likely, mutually exclusive and exhaustive cases, m cases are favorable to the A event and n are not favorable to the event A, m : n is called as odds in favour of A, n : m is called as odds against the event A.

 

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