Probability proof - what formulas are needed here?

[SavvyAA3]

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?

 

[exk]

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).

 

[SavvyAA3]

Please could you show me the steps you would take

 

[Zone Ranger]

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
P(A|B&#039;)&gt;P(A)
then

\frac{P(A\cap B&#039;)}{P(B&#039;)}&gt;P(A)

\frac{P(B&#039;|A)P(A)}{P(B&#039;)}&gt;P(A)

\frac{P(B&#039;|A)}{P(B&#039;)}&gt;1

P(B&#039;|A)&gt;P(B&#039;)

1-P(B&#039;|A)&lt;1-P(B&#039;)

P(B|A)&lt;P(B)

the other one isn't much different

 

[SavvyAA3]

Thanks soo much!