It doesn’t seem to be true that P(A and B) for two events happening consecutively is the same as P(A and B) for two events happening simultaneously.
So we should never use the notation P(A ∩ B), the probability of the intersection of A and B for the first case? (I have been using that notation whenever seeing A and B)
For example, let A be the event you roll an even # on a die, let B be the event you roll a 2.
P(A and B), the probability of A happening and then B happening is (1/2)(1/6) = 1/12 by the rule for independent events..
But P(A∩B), the probability of the events happening simultaneously is 1/6. There is only one case out of 6 where you would roll a 2 and even.
So P(A and B) ≠ P(A ∩ B) here.. what’s wrong??
April 27th 2011, 05:13 PM
You are confusing yourself.
Roll One Die Twice: P(even) = 1/2, then p(2) = 1/6, p(even and 2) = (1/2)*(1/6) = 1/12
Roll Two Dice Simultaneously: P(#1 is even) = 1/2, and p(#2 is 2) = 1/6, then p(#1 is even and #2 is 2) = (1/2)*(1/6) = 1/12
Roll One Die Once: P(even and 2) = p(2) = 1/6 -- These two events are not independent with only one roll of one die.
April 27th 2011, 09:04 PM
Thank you, can P(A and B) , P(A ∩ B) be used interchangeably? Or do they mean something different?
April 28th 2011, 03:51 AM
I cannot think of a usage where they actually differ. Perhaps if you were emphasizing logical structure or set structure...