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You throw two dice.

a) What is the probability that the difference in the values showing equals 1? (4,3 and 3,4 both qualify)

b) What is the covariance between the value of the first die and the value of the second die?

Attempted Solutions:

a) I got .2777 because there are 36 elements in the event space and 10 of which have a difference of 1.

b) I know that $Cov(X,Y) = E(XY) - E(X)E(Y)$. $E(X)$ and $E(Y)$ are both 3.5. I am not sure how to obtain $E(XY)$ though.

1 Answers 1

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a) Correct.

b) Hint: Independence.

Also $\mathsf E(XY) =\sum\limits_{x=1}^6\sum\limits_{y=1}^6 \dfrac{xy}{36}$

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    I got a covariance of 0. Is that right? – 2017-02-08
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    Also, I wasn't sure if the question was asking about the covariance for having a difference of 1 or not. In that case, wouldn't it be 0 as well because no matter what you roll first, there is always a 1/6 chance that the difference will be 1? – 2017-02-08
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    Actually, if you roll $2$ first there is a $1/3$ chance to have a difference of $1.$ That's how you got a value greater than $1/6$ for part a). But the difference of the dice is neither "the value of the first die" nor "the value of the second die," so it seems not to be relevant to the covariance question. – 2017-02-08
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    Oh, right. Forgot about that. – 2017-02-08