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(D) The total number of ways to choose two distinct numbers $a$ and $b$ from the set $\{1, 2, 3, \dots, 30\}$ is given by ${}^{30}C_2 = \frac{30 	ime

Vedclass · Accepted Answer

$\frac{47}{87}$

Vedclass · Answer

(D) The total number of ways to choose two distinct numbers $a$ and $b$ from the set $\{1, 2, 3, \dots, 30\}$ is given by ${}^{30}C_2 = \frac{30 	imes 29}{2} = 435$. For $a^2 - b^2$ to be divisible by $3$,we consider the numbers modulo $3$. Let $S_0 = \{3, 6, \dots, 30\}$ (numbers divisible by $3$,count $= 10$),$S_1 = \{1, 4, \dots, 28\}$ (numbers $\equiv 1 \pmod 3$,count $= 10$),and $S_2 = \{2, 5, \dots, 29\}$ (numbers $\equiv 2 \pmod 3$,count $= 10$). $a^2 - b^2 \equiv 0 \pmod 3$ implies $a^2 \equiv b^2 \pmod 3$. If $x \in S_0$,$x^2 \equiv 0 \pmod 3$. If $x \in S_1 \cup S_2$,$x^2 \equiv 1 \pmod 3$. Case $1$: Both $a, b \in S_0$. Number of ways $= {}^{10}C_2 = 45$. Case $2$: Both $a, b \in S_1 \cup S_2$. Number of ways $= {}^{20}C_2 = 190$. Total favorable cases $= 45 + 190 = 235$. Required probability $= \frac{235}{435} = \frac{47}{87}$.

Two numbers $a$ and $b$ are chosen at random from the set of the first $30$ natural numbers. The probability that $a^2 - b^2$ is divisible by $3$ is

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