probability of independent events

Apr 2014
275
1
canada
i have drawn the venn diagram for the question... i am not sure it's correct or not. the question states both are independent events. i dont understand the working in part b ... for me , my ans would be p(A) +p(B).. correct me if i am wrong. thanks in advance. DSC_0019.JPGDSC_0020.JPGDSC_0020.JPG
 
Mar 2012
117
21
malaysia
i have drawn the venn diagram for the question... i am not sure it's correct or not. the question states both are independent events. i dont understand the working in part b ... for me , my ans would be p(A) +p(B).. correct me if i am wrong. thanks in advance. View attachment 31189View attachment 31190View attachment 31190
you have to minus the probability of intersection of set A and B. the way i understand it: for example, if A is {1,2,3} and B is {3,4,5}, AUB is {1,2,3,4,5}, not {1,2,3,3,4,5}. if you added probability of obtaning A and probability of obtaining B, and did not minus the intersection, you would have calculated the probability of obtaining 3 twice, that's why u have to minus the probability of the intersection ( the probability of obtaining a 3)
 
Last edited:
Mar 2012
117
21
malaysia
take note that when A and B are mutually exclusive, then P(AUB)=P(A)+P(B). when they are mutually exclusive, none of the elements of A and B are the same. In one trial you either get something from A or from B, you cannot get a result that is from both A and B like the 3 just now. if they are mutually exclusive, something like "A is {1,2,3} and B is {3,4,5}" is impossible. Possible scenario is that it will be like "A is {1,2,3} and B is {4,5,6}".

when it is not mentioned to be mutually exclusive, you cannot assume that it is. like in this question
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Another way of looking at it: Imagine there are 1000 items the 1/20 of 1000= 50 will have defect A and 1/10 of 1000= 100 will have defect B. Since the two defects are independent (1/20)(1/10)= 1/200 of 1000= 5 will have both defects. So 50- 5= 45 will have only defect A, 100- 5= 95 will have only, 5 will have both defects and 1000- 45- 95- 5= 855 will have no defects.

Thus the probability of both defects is 5/1000= .005, probability of defect A only is 45/1000= .045, probability of defect B only= 95/1000= .095, and probability of no defects is 855/1000= .855.
 
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