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Couple things to remember for this proof: | Couple things to remember for this proof: | ||
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| + | <MATH>P(A|B)= P(A \bigcap B)/P(B) </MATH> | ||
| + | <br> | ||
| + | <br> | ||
| + | so | ||
| + | <br> | ||
| + | <br> | ||
| + | <MATH>P(A \bigcap B)= P(A|B)*P(B) </MATH> | ||
| + | <br> | ||
| + | <br> | ||
| + | Make sure you use the theorem of total probability, which states: | ||
| + | <br> | ||
| + | <br> | ||
| + | <MATH>P(A)=P(A|C)P(C)+P(A|C^c)P(C^c) \!</MATH> | ||
| + | <br> | ||
| + | <br> | ||
| + | Try and rearrange what we want to proof so it looks like what we know is true. | ||
Latest revision as of 10:12, 17 September 2008
Couple things to remember for this proof:
$ P(A|B)= P(A \bigcap B)/P(B) $
so
$ P(A \bigcap B)= P(A|B)*P(B) $
Make sure you use the theorem of total probability, which states:
$ P(A)=P(A|C)P(C)+P(A|C^c)P(C^c) \! $
Try and rearrange what we want to proof so it looks like what we know is true.
