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(D) Total ways to choose two distinct numbers from $30$ is $^{30}C_{2} = \frac{30 	imes 29}{2} = 435$. For $a^2 - b^2$ to be divisible by $3$,$(a-b)(

Vedclass · Accepted Answer

$47$

Vedclass · Answer

(D) Total ways to choose two distinct numbers from $30$ is $^{30}C_{2} = \frac{30 	imes 29}{2} = 435$. For $a^2 - b^2$ to be divisible by $3$,$(a-b)(a+b)$ must be divisible by $3$. This happens if $a \equiv b \pmod{3}$ or $a \equiv -b \pmod{3}$. Let the sets based on remainders modulo $3$ be: $D_{1} = \{1, 4, 7, \dots, 28\}$ (size $10$,remainder $1$) $D_{2} = \{2, 5, 8, \dots, 29\}$ (size $10$,remainder $2$) $D_{3} = \{3, 6, 9, \dots, 30\}$ (size $10$,remainder $0$) Favorable cases: $1$. Both $a, b$ from the same set: $^{10}C_{2} + ^{10}C_{2} + ^{10}C_{2} = 45 + 45 + 45 = 135$. $2$. One from $D_{1}$ and one from $D_{2}$: $^{10}C_{1} 	imes ^{10}C_{1} = 100$. Total favorable cases $= 135 + 100 = 235$. Probability $P = \frac{235}{435} = \frac{47}{87}$.

Two numbers $a$ and $b$ are chosen at random from the set of first $30$ natural numbers. Then the probability that $a^2 - b^2$ is divisible by $3$ is (in $/87$)

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