chapter 4 additional notes and examples

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Please note for conditional probabilities I gave you 2 formulas. One deals with probabilities of events, the other deals with numbers in an event. ) ( () ( PB PA (Α ∩ |Α)= Ν(Α ∩Β) Ν(Α) This 2 nd formula can be useful for some problems. It is particularly helpful for combination problems. HOWEVER, AN ALTERNATIVE APPROACH USES LOGIC. Example: Problem set problem 4a) The question asks what is the conditional probability of getting a black ball on the second draw given the first ball is black. Since there are only 2 black balls, there is only 1 way to get a black ball on the 2 nd draw given the first draw was black. Also, since we already drew one ball there are only 4 balls left. Therefore, our conditional probability is ¼. Remark: This approach works nicely in cases of drawing without replacement (whether that is balls from a box or letters from a string). Note: If we wanted the probability of 1 st draw black and 2 nd draw black; we would use the total probability rule.

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