ku Posted March 19, 2005 Share Posted March 19, 2005 Toss a die twice and take the sums. Let A be the event that the sum of the two numbers rolled is 6. Let B be the event that you roll a 6 on the first roll. Is there a negative relationship between A and B? I believe there is, but I'd like to have my working checked out. [math]AB = \emptyset[/math] and therefore [math]P(AB)=0[/math]. [math]A = \{(1,5), (2,4), (3,3), (4,2), (1,5)\}[/math] and therefore [math]P(A)=\frac{5}{36}[/math] [math]B = \{(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}[/math] and therefore [math]P(B)=\frac{6}{36}=\frac{1}{6}[/math] [math]P(AB) < P(A)P(B)[/math] [math]\frac{P(AB)}{P(B)}=P(A)[/math] [math]P(A|B) < P(A)[/math] Link to comment Share on other sites More sharing options...
matt grime Posted March 19, 2005 Share Posted March 19, 2005 What on earth is a negative relation? Link to comment Share on other sites More sharing options...
ku Posted March 20, 2005 Author Share Posted March 20, 2005 According to my notes, when [math]P(A|B) < P(A)[/math] then there is a negative relationship between A and B. The probability of A happening given that B happens is less than the probability that A happens by itself, which means that B is influencing A. Link to comment Share on other sites More sharing options...
dan19_83 Posted March 21, 2005 Share Posted March 21, 2005 But if B happens, doesn't that mean it is impossible for A to happen? you've already thrown 6 so another throw will give a sum greater than 6. Link to comment Share on other sites More sharing options...
matt grime Posted March 21, 2005 Share Posted March 21, 2005 So, have you worked out P(A|B)? As the above post indicates, it's straightforward. Link to comment Share on other sites More sharing options...
dan19_83 Posted March 21, 2005 Share Posted March 21, 2005 P(A|B) is the probability of A (sum of two throws = 6) given that B (a 6 was thrown in the first throw) has already occured. So if you throw for a second time then the value of the two throws will always be bigger than six. therefore the probability is zero. Link to comment Share on other sites More sharing options...
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