Understanding Joint Probability with No-Replacement Rule

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Joint probability P(A.B) generally does not require concern for the order of events A and B. However, in scenarios involving no replacement, such as drawing balls from a set, the order becomes significant. For example, when calculating P(R|B), the probability of drawing a red ball given a blue ball was previously drawn, the sequence of draws matters. Despite this, the joint probability P(RB) can still be calculated without regard to order, provided the no-replacement condition is factored into the definitions of the events. Understanding these nuances is crucial for accurately applying joint and conditional probabilities in practical situations.
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If A and B are two events and I want to look at their joint probability P(A.B) do I have to be concerned with the order in with A and B occur?

Saurav
 
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No, in general. However, consider:

A = {accumulation > x}

B = {accumulation < x}.

In this case, there is a natural temporal ordering of the events A and B; B almost certainly occurs before A.
 
Thanks EnumaElish

So now if I am looking at a problem where I have 3 red balls and 4 blue balls and if I pickup two balls want to find the probability P(R|B) (i.e prob of picking Red ball given Blue ball was already picked). In this conditional probability case, the order in which the balls were picked is important, am i correct? However the joint probability P(RB) = P(R|B).P(B) = P(B|R).P(R) is not concerned with the order in which the balls were picked?

Saurav
 
You are correct that as long as you are not replacing each draw (by putting a drawn ball back into the bin), the order of draws will matter. This does not have to conflict with the definition of joint probability using conditional probabilities as long as you define each event by taking the no-replacement rule into account. So if you define P(R|B) as the probability of drawing a red having drawn a blue, then P(RB) will be defined accordingly.
 
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