Two integers $x$ and $y$ are chosen with replacement from the set $\{0, 1, 2, 3, \ldots, 10\}$. Then the…
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:
Solution
Given set is $\{0,1,2,3,...,10\}$.
So, total number of outcomes are $11 \times 11 = 121$.
Favourable outcomes are:
$\{0,6\}, \{0,7\}, \{0,8\}, \{0,9\}, \{0,10\}, \{1,7\}, \{1,8\}, \{1,9\}, \{1,10\}, \{2,8\}, \{2,9\}, \{2,10\}, \{3,9\}, \{3,10\}, \{4,10\}$.
And the same number of elements when $x$ and $y$ are interchanged.
So, the number of favourable outcomes are $30$.
So, the required probability is given by $P(E) = \frac{30}{121}$.