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Math Help - Two proofs (probability)

  1. #1
    Tau
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    Two proofs (probability)

    Hello

    Help me prove these two classical rules

    1. A and B are two events. Show that the probability that exactly one of these events will occur is P(A) + P(B) - 2P(A \cap B)

    2. We know that P(A \cup B) = P(A) + P(B) - P(A \cap B).
    Prove that P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)

    Thank you.
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    Quote Originally Posted by Tau View Post
    1. A and B are two events. Show that the probability that exactly one of these events will occur is P(A) + P(B) - 2P(A \cap B)

    2. We know that P(A \cup B) = P(A) + P(B) - P(A \cap B).
    Prove that P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)
    Exactly one is: P(A\cap B^c)+P(B\cap A^c).
    Recall that P(A\cap B^c)=P(A)-P(A\cap B).

    In #2 let D=A\cup B. The expand P(D\cup C).
    The go back and make subsitutions.
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  3. #3
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    Hello, Tau!

    2. We know that P(A \cup B) = P(A) + P(B) - P(A \cap B).


    Prove that: . P(A \cup B \cup C) \; = \;\begin{Bmatrix} P(A) + P(B) + P(C) \\<br />
- P(A \cap B) - P(A \cap C) - P(B \cap C) \\ + P(A \cap B \cap C) \end{Bmatrix}

    P(A \cup B \cup C)

    . . =\qquad\qquad\qquad\qquad P\bigg[(A \cup B) \cup C\bigg]

    . . =\qquad\qquad P(A \cup B) + P(C) - P\bigg[ (A \cup B) \cap C\bigg]

    . . =\qquad\qquad P(A \cup B) + P(C) - \underbrace{P\bigg[(A \cap C) \cup (B \cap C)\bigg]}

    . . =\qquad P(A \cup B) + P(C) - \overbrace{\bigg[P(A\cap C) + P(B\cap C) - P([A \cap C] \cap [B \cap C])\bigg]}


    . . =\qquad \underbrace{P(A \cup B)} + P(C) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)

    . . =\;\overbrace{P(A) + P(B) - P(A \cap B)} + P(C) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)


    . . =\; P(A) + P(B) + P(C) - P(A\cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)


    . . . . . . . . . . . . ta-DAA!

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