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Conditional probability of cows

1) You live on a farm and have six cows: two of them are black, three are white and one is black and white (that is, black on one side and white on the other). If you look outside the window and see two black cows, what's the probability that one of them is your black and white cow?

Attempt:

P(black and white cow| you see 2 black cows) = P(black and white cow| one is completely black and you see 2 black cows) P(one is completely black | you see 2 black cows) + P(black and white cow| one is not completely black and you see 2 black cows) P(one is not completely black| you see 2 black cows).

I got the above by simply using the rule: P(E|F) = P(E|GF)P(G|F) + P(E|$G^c$F)P($G^c$|F). I get the correct answer using this method, but I asked my professor today and he said I can't do this because one of my events are not disjoint. I thought intially I had made an error, but when I look back over it, I am only introducing one another event: G, and then I only subsequently work with the complement of G. Am I correct?

2) Not in relation to the question above, but when I have a joint probability density $f(x,y)$ and I want to determine if $X$ and $Y$ are independent, is to okay to say they are independent provided I can write $f(x,y) = f_X(x) f_Y(y)$ and that the limits of $x, y$ do not depend on each other?

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CAF
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  • Exactly what do you mean by one is completely black and one is not completely black? For the first do you mean at least one, or exactly one? Whichever is meant, one is not completely black is not its negation, so you are not describing complementary events. – Brian M. Scott Dec 11 '12 at 15:17
  • I meant exactly one. What would the negation/complement be? – CAF Dec 11 '12 at 15:25
  • It is not the case that exactly one of the cows is completely black, i.e., either both cows are completely black, or neither cow is completely black. – Brian M. Scott Dec 11 '12 at 15:41
  • there is intersection since the possible cases are: $c_1 = (B,B)$, $c_2 = (B,W)$, $c_3 = (W, W)$, $c_4 = (B, B/W)$ and $c_5 = (W, B/W)$. When you say "one is completely black", then you have the set of cases ${c_1, c_2, c_4}$, while when you say "one is not completely black, then you have ${c_4, c_5}$. So the intersection is ${c_4}$, which is not empty and hence the event are not complementary – the_candyman Dec 11 '12 at 15:43
  • @the_candyman: So that means in general, $G^cG = $empty set?@Brian: If I take the latter of your two suggestions, and sub it in, I do indeed still get the correct answer. Does that mean it is all good now? I am not sure if I can use this since it is given that there are only 1 B/W cow. If I take the former of your suggestions, I get the correct answer, with no problems. – CAF Dec 11 '12 at 16:18
  • What about my other question about the independence of RV's? – CAF Dec 11 '12 at 16:28
  • For your second question, the answer is yes. In fact you technically don't even need to add the "limits don't depend on each other" (but informally you should). The reason you don't really need it is that joint densities, densities are really defined respectively over all $(x,y)$, and over all $x$, all $y$. We just get accustomed to not paying attention to the places where the density is $0$. – André Nicolas Dec 11 '12 at 16:58
  • @CAF: I don't understand the meaning of this $G^cG$. – the_candyman Dec 12 '12 at 12:38
  • @CAF: $A = {c_1, c_2, c_3, c_4, c_5}$ is the set of all possible cases. If $G \subset A$ is the event "one is completely black", then you have 3 cases, which are ${c_1, c_2, c_4}$. So the complementary set is $ G^{c} = A \setminus G = {c_3, c_5}$. This set can be read as "one is white and the other one is not black (so the latter can be white or half white/half black)" which is different from "one is not completely black". This event, say $H$, is $H = {c_4, c_5}$. Hope this help – the_candyman Dec 12 '12 at 12:45
  • @CAF: finally, the definition of complementary set $G^c$ with respect to the "universe" set $A$ is that: $G^c = A \setminus G$ and hence we have that $G \cap G^c = {}$ and $G \cup G^c = A$. – the_candyman Dec 12 '12 at 12:49
  • Thanks the_candyman, my book uses $GG^c$ to mean the intersection of $G$ and $G^c$. I never understood it first because it seemed to mean the sets were being multiplied, which makes no sense. – CAF Dec 12 '12 at 16:51
  • Well, in a certain way that's not really a non-sense. In fact, on some book about Boole's algebra, the "AND" operator is replaced by a multiplication and set intersection is some how equivalent to the "AND". – the_candyman Dec 13 '12 at 12:13

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