Solution: 2022 Winter Final - 21

We need to determine whether $X$ and $Y$ are independent or dependent.

Definition of Independence of Random Variables

Two random variables $X$ and $Y$ are independent if: $P(X = x | Y = y) = P(X = x) \cdot P(Y = y)$ for all $x$ and $y$. If this condition does not hold, they are dependent.

Relationship Between X and Y

What do the variables represent?

• $X = \max(D_1, D_2)$: The maximum value observed from two six-sided dice rolls.
• $Y = D_1 + D_2$: The sum of the two dice rolls.

Key Observation

The maximum value of two dice rolls is inherently tied to their sum:

For example, if $Y = 12$ (the sum is maximized), it means both dice rolled a 6, and therefore, $X = 6$.

Certain values of $Y$ restrict the possible values of $X$. This means knowing $Y$ influences our knowledge of $X$.

Counter-Example for Independence

To prove that $X$ and $Y$ are dependent, we need to find a counter-example where the independence condition does not hold for a single pair of $x$ and $y$ values.

Let’s consider the following:

• $X = 6$ (Maximum of two rolls is 6)
• $Y = 7$ (Sum of two rolls is 7)

We will calculate:

1. $ P(X = 6 \land Y = 7) $
2. $ P(X = 6) $
3. $ P(Y = 7) $
4. Check if the independence condition holds.

Calculate Each Probability

Possible pairs $(D_1, D_2)$ such that $ Y = 7 $: Possible pairs are: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$

Among these pairs, the ones where the maximum value is 6 are: $ (1,6), (6,1) $

There are 2 favorable outcomes out of $ 36 $ total outcomes (since two six-sided dice have $ 6 \times 6 = 36 $ total possibilities). So: $ P(X = 6 \land Y = 7) = \frac{2}{36} = \frac{1}{18} $

For $ X = 6 $, the possible pairs are: $(1,6), (2,6), (3,6), (4,6), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$

There are 11 favorable outcomes out of $ 36 $ total outcomes. So: $P(X = 6) = \frac{11}{36}$

The possible pairs for $ Y = 7 $ are: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$

There are 6 favorable outcomes out of $ 36 $ total outcomes. So: $P(Y = 7) = \frac{6}{36} = \frac{1}{6}$

Compare the Independence Condition

Substituting into the independence formula:

$ P(X = 6 \land Y = 7) \stackrel{?}{=} P(X = 6) \cdot P(Y = 7) $

Substituting the values:

$ \frac{1}{18} \stackrel{?}{=} \frac{11}{36} \cdot \frac{1}{6} $

Simplify the right-hand side:

$ \frac{11}{216} $

Convert the left-hand side to have the same denominator:

$ \frac{1}{18} = \frac{12}{216} $

Clearly:

$ \frac{12}{216} \neq \frac{11}{216} $

The equality does not hold, violating the independence condition. This proves that $X$ and $Y$ are not independent.