Negative Relationships

[ku]

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]

[matt grime]

What on earth is a negative relation?

[ku]

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.

[dan19_83]

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.

[matt grime]

So, have you worked out P(A|B)? As the above post indicates, it's straightforward.

[dan19_83]

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.