: This was given as an examination question for Year II Probability
students.
: Something must be wrong. The first part of the question said to state the
: Axioms of Probability. So, is there a connection?
:
:
: ----- Original Message -----
: From: <[log in to unmask]>
: To: Isaac Dialsingh <[log in to unmask]>
: Sent: Monday, May 20, 2002 6:07 AM
: Subject: Re: Proof
:
:
: :
: : Dear Isaac,
: :
: : Since this is a fairly standard result , namely P(A or B) = P(A) +
P(B) -
: : P(A and B) for independent events, what exactly is the problem ?
: :
: :
: : =========
: :
: : Best wishes
: :
: : Paul Garcia
: : Senior Property Review Officer (Demographics)
: : 25221
: : =====================================
: : Hoc fama mihi cursum sinuosum secuta nuntiat
: :
: :
: :
: : Isaac
: : Dialsingh To: [log in to unmask]
: : <[log in to unmask] cc:
: : NET.TT> Subject: Proof
: : Sent by: A
: : UK-based
: : worldwide
: : e-mail
: : broadcast
: : system mailing
: : list
: : <allstat@JISCM
: : AIL.AC.UK>
: :
: :
: : 19/05/2002
: : 19:39
: : Please respond
: : to Isaac
: : Dialsingh
: :
: :
: :
: :
: :
: :
: : Can someone explain why this proof is incorrect? In addition, what must
we
: : assume for this to be correct?
: :
: : Let S be the sample space of the experiment and let E be any event in S.
: : Denote the number of elements in an event E by n(E) and let the number
of
: : elements in S be N, then
: :
: : P(E) = n(E)/N
: :
: : Let X = A intersect B'
: : Y = A intersect B
: : Z = A' intersect B. Then
: : A union B = (A intersect B') union (A intersect B) union (A' intersect
B)
: : and the events
: : (A intersect B') , (A intersect B) and (A' intersect B) are
: : disjoint. hence,
: :
: : P(A union B) = n(A union B)/N
: : = (n(X) + n(Y) + n(Z)) / N
: : = [ [n(X) + n(Y)] + [n(Y) + n(Z)] - n(Y) ] / N
: : = [ n(A) + n(B) -n(A intersect B) ] / N
: : = n(A)/N + n(B)/N + n(A intersect B)/N
: : = P(A) + P(B) - P(A intersect B)
: :
: :
: :
: :
: :
: :
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