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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 	imes

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 ${}^{30}C_2 = \frac{30 	imes 29}{2} = 435$.We know that $a^2 - b^2$ is divisible by $3$ if $a^2 \equiv b^2 \pmod{3}$.For any integer $n$, $n^2 \pmod{3}$ can be $0$ (if $n$ is a multiple of $3$) or $1$ (if $n$ is not a multiple of $3$).Let $S_0$ be the set of numbers divisible by $3$ $(|S_0| = 10)$ and $S_1$ be the set of numbers not divisible by $3$ $(|S_1| = 20)$.$a^2 - b^2$ is divisible by $3$ in two cases:Case $1$: Both $a$ and $b$ are in $S_0$. The number of ways is ${}^{10}C_2 = \frac{10 	imes 9}{2} = 45$.Case $2$: Both $a$ and $b$ are in $S_1$. The number of ways is ${}^{20}C_2 = \frac{20 	imes 19}{2} = 190$.Total favorable cases $= 45 + 190 = 235$.Therefore, the required probability $= \frac{235}{435} = \frac{47}{87}$.

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

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