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(A) The total number of ways to choose two integers $x$ and $y$ with replacement from the set $\{0, 1, 2, \ldots, 10\}$ is $11 	imes 11 = 121$.We wan

Vedclass · Accepted Answer

$\frac{30}{121}$

Vedclass · Answer

(A) The total number of ways to choose two integers $x$ and $y$ with replacement from the set $\{0, 1, 2, \ldots, 10\}$ is $11 	imes 11 = 121$.We want to find the number of pairs $(x, y)$ such that $|x-y| > 5$.Case $1$: $x - y > 5 \implies x - y \in \{6, 7, 8, 9, 10\}$.If $x=6, y=0$; if $x=7, y=0, 1$; if $x=8, y=0, 1, 2$; if $x=9, y=0, 1, 2, 3$; if $x=10, y=0, 1, 2, 3, 4$.The number of such pairs is $1 + 2 + 3 + 4 + 5 = 15$.Case $2$: $y - x > 5 \implies y - x \in \{6, 7, 8, 9, 10\}$.By symmetry,the number of such pairs is also $15$.Total favorable outcomes $= 15 + 15 = 30$.Therefore,the required probability is $\frac{30}{121}$.

Two integers $x$ and $y$ are chosen with replacement from the set $\{0, 1, 2, 3, \ldots, 10\}$. Then the probability that $|x-y| > 5$ is:

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