Probability proof - what formulas are needed here?

  • Thread starter SavvyAA3
  • Start date
23
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Main Question or Discussion Point

If events A and B are in the same sample space:
  • .
Proove that if P(A I B') > P(A) then P(B I A) < P(B)

(where B' is the Probability of A given not B)


  • .
Proove that if P(A I B) = P(A) then P(B I A) = P(B)

do we assume independence here so that P(A I B) = [P(A)*P(B)]/ P(B) = P(A) and state that since P(A n B) = P(B n A) that P(B I A) = [P(B)*P(A)] / P(A) = P(B) or is it wrong to assume independence here?
 

Answers and Replies

exk
119
0
For the second proof use the fact that P(A|B)=P(A&B)/P(B) and similarly for the other one. You can't assume independence, but it is easy to see that they are using the usual definition of independence P(A&B)=P(A)P(B).
 
23
0
Please could you show me the steps you would take
 
If events A and B are in the same sample space:
  • .
Proove that if P(A I B') > P(A) then P(B I A) < P(B)

(where B' is the Probability of A given not B)

...
assume
[tex]P(A|B')>P(A)[/tex]
then

[tex]\frac{P(A\cap B')}{P(B')}>P(A)[/tex]

[tex]\frac{P(B'|A)P(A)}{P(B')}>P(A)[/tex]

[tex]\frac{P(B'|A)}{P(B')}>1[/tex]

[tex]P(B'|A)>P(B')[/tex]

[tex]1-P(B'|A)<1-P(B')[/tex]

[tex]P(B|A)<P(B)[/tex]

the other one isn't much different
 
23
0
Thanks soo much!!
 

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