Solution: 2022 Winter Final - 22

Author: Michiel Smid

Question

Let $D_1$ and $D_2$ be the result of tossing a $6$-sided die twice. Define the random variables \[ X=2\cdot D_1 \] and \[ Y= \begin{cases} 1 & \text{if $D_1+D_2=7$} \\ 0 & \text{otherwise.} \enspace . \end{cases} \] Which of the following is true?

(a)

None of the above

(b)

$X$ and $Y$ are independent

(c)

$X$ and $Y$ are not independent

Solution

Technically, to solve this question, we would have to prove that for every possible real number, that $Pr(X = x \cap Y = y) = Pr(X = x) \cdot Pr(Y = y)$. \ \

Here is more of a conceptual reason why the two random variables are independent. \

The event Y, does not depend on the results of event X. Since the probability of