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Choose one of the points $(0,0)$, $(0,1)$, $(1,0)$ and $(1,1)$ at random. Let $X$ be the first coordinate and $Y$ the second. Are $X$ and $Y$ independent?

They would be independent since event $Y$ does not alter what event $X$ would be. Since $X$ could be either $0$ or $1$, regardless of what $Y$ is.

Is this correct?

3 Answers 3

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Yes, that is correct.

A little more technical: When you say that one of these four are chosen at random, I assume you mean that each of these 4 outcomes has a probability of $\frac{1}{4}$. And, as such, the probability of $Y$ being a $0$ (or $1$) does not differ depending on whether $X$ is a $0$ or $1$

Mathematically:

$P(Y=0|X=0) = P(Y=0|X=1)$ (= $\frac{1}{2}$) and

$P(Y=1|X=0) = P(Y=1|X=1)$ (= $\frac{1}{2}$)

Even more general, events $A$ and $B$ are independent if and only if $P(A) = P(A|B)$. And indeed we have:

$P(Y=0) = P(Y=0|X=0)$ (= $\frac{1}{2}$) and

$P(Y=0) = P(Y=0|X=1)$ (= $\frac{1}{2}$) and

$P(Y=1) = P(Y=1|X=0)$ (= $\frac{1}{2}$) and

$P(Y=1) = P(Y=1|X=1)$ (= $\frac{1}{2}$)

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    " events $A$ and $B$ are independent if and only if $P(A)=P(A\mid B)$". Not so. Events $A$ and $B$ are said to be independent if $P(A \cap B)=P(A)P(B)$ and since this is a definition, you can replace "if" by "if and only if" if you like. This definition holds even if $P(B)=0$ in which case the $P(A\mid B)$ in your definition is undefined. – 2017-02-23
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    @DilipSarwate You are correct! I should have just said 'if'. I'll correct that, thank you! – 2017-02-23
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Yes you are right. Event Y not depends on X. We don't have any condition.

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Nope. They can be independent if 'choose one of four points at random' means 'with uniform distribution' or, equivalently, 'with equal probability'.

However, although that is quite reasonable assumption, it is not defined in the problem, so we might suppose a non-uniform distribution. And for probabilities, say, $$P((0,0))=P((1,1))=4/10$$ and $$P((0,1))=P((1,0))=1/10$$ we would get a significant correlation between $X$ and $Y$.