tag:blogger.com,1999:blog-4631023797563841554.post3784505633251479373..comments2020-07-20T13:24:50.941-05:00Comments on Maverick Christian: “If A, then probably C” entails “Probably, if A then C”Maverick Christianhttp://www.blogger.com/profile/04286456663634536819noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-4631023797563841554.post-73487969610526514852014-02-15T08:41:10.236-06:002014-02-15T08:41:10.236-06:00Ack! Use unicode man! ;-) Reformatting it: 1) P(...Ack! Use unicode man! ;-)<br /><br />Reformatting it:<br /><br />1) P(A|B) ≤ 1<br />⇓<br />2) P(A|B)(1-P(A)) ≤ (1-P(A))<br />⇓<br />3) P(A|B) ≤ P(A|B)P(A)+1-P(A)<br />⇓<br />4) P(A|B) ≤ P(A|B)P(A)+1-P(A)<br />⇓<br />5) P(A|B) ≤ 1-(P(A)-P(A∩B))<br /><br />I&#39;m only a math minor, but there seems there might be a problem between (4) and (5). Note that while this is true:<br /><br />P(A|B)P(B) = P(A∩B)<br /><br />The following is not true in general:<br /><br />P(A|B)P(A) = P(A∩B)<br /><br />Example: There is a sack containing balls numbered 1, 2, 4, and 6. Let P(A) represent the probability that I have drawn an even numbered ball, which in this case P(A) = 0.75. Let B represent the outcome that I had drawn a ball less than 4. Then P(A|B) represents the probability that I have drawn an even numbered ball given that I had drawn a ball less than 4, and P(A|B) = 0.5. Yet the following equation is not true:<br /><br />P(A|B)P(A) = P(A∩B)<br /><br />For P(A∩B) = 50% yet P(A|B)P(A) = 37.5%Maverick Christianhttps://www.blogger.com/profile/04286456663634536819noreply@blogger.comtag:blogger.com,1999:blog-4631023797563841554.post-43879000016376580262013-11-30T20:31:30.817-06:002013-11-30T20:31:30.817-06:00There is a much simpler way to do this P(A|B)&lt;=...There is a much simpler way to do this<br />P(A|B)&lt;=1<br />P(A|B)(1-P(A))&lt;=(1-P(A))<br />P(A|B)&lt;=P(A|B)P(A)+1-P(A)=1-(P(A)-P(AnB))=1-P(An~B)=P(~AuB)=P(A--&gt;B).Butch Crantonnoreply@blogger.com