If two different numbers are taken from the s | vedclass

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(B) Let the set be $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.Total number of ways to choose two distinct numbers is $n(S) = \binom{11}{2} = \frac{11

Vedclass · Accepted Answer

$\frac{6}{55}$

Vedclass · Answer

(B) Let the set be $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.Total number of ways to choose two distinct numbers is $n(S) = \binom{11}{2} = \frac{11 	imes 10}{2} = 55$.Let the two numbers be $x$ and $y$ where $x > y$.We require $(x+y)$ to be a multiple of $4$ and $(x-y)$ to be a multiple of $4$.If $(x+y)$ and $(x-y)$ are both multiples of $4$,then their sum $(x+y) + (x-y) = 2x$ must be a multiple of $4$,implying $x$ is an even number.Similarly,their difference $(x+y) - (x-y) = 2y$ must be a multiple of $4$,implying $y$ is an even number.Thus,both $x$ and $y$ must be even numbers from the set $\{0, 2, 4, 6, 8, 10\}$.Let the chosen numbers be $x, y \in \{0, 2, 4, 6, 8, 10\}$.For $(x+y)$ and $(x-y)$ to be multiples of $4$,$x+y \equiv 0 \pmod{4}$ and $x-y \equiv 0 \pmod{4}$.This implies $x \equiv y \pmod{4}$.Possible pairs $(x, y)$ with $x > y$ such that $x \equiv y \pmod{4}$:If $x, y \equiv 0 \pmod{4}$: $(4, 0), (8, 0), (8, 4)$.If $x, y \equiv 2 \pmod{4}$: $(6, 2), (10, 2), (10, 6)$.Total favorable outcomes $n(E) = 3 + 3 = 6$.Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{55}$.

If two different numbers are taken from the set $\{0, 1, 2, 3, \dots, 10\}$,then the probability that their sum as well as their absolute difference are both multiples of $4$ is:

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