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(D) The total number of ways to choose $3$ integers from $20$ is ${}^{20}C_3 = \frac{20 	imes 19 	imes 18}{3 	imes 2 	imes 1} = 1140$. We categori

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$\frac{32}{85}$

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(D) The total number of ways to choose $3$ integers from $20$ is ${}^{20}C_3 = \frac{20 	imes 19 	imes 18}{3 	imes 2 	imes 1} = 1140$. We categorize the numbers from $1$ to $20$ based on their remainder when divided by $3$: $R_0 = \{3, 6, 9, 12, 15, 18\}$ (count $6$) $R_1 = \{1, 4, 7, 10, 13, 16, 19\}$ (count $7$) $R_2 = \{2, 5, 8, 11, 14, 17, 20\}$ (count $7$) The sum is divisible by $3$ in the following cases: $(I)$ All three numbers from $R_0$: ${}^{6}C_3 = 20$. $(II)$ All three numbers from $R_1$: ${}^{7}C_3 = 35$. $(III)$ All three numbers from $R_2$: ${}^{7}C_3 = 35$. $(IV)$ One number from each set $R_0, R_1, R_2$: ${}^{6}C_1 	imes {}^{7}C_1 	imes {}^{7}C_1 = 6 	imes 7 	imes 7 = 294$. Total favorable outcomes $= 20 + 35 + 35 + 294 = 384$. Probability $= \frac{384}{1140} = \frac{32}{85}$.

Three numbers are chosen at random from $1$ to $20$. The probability that the sum of the three numbers is divisible by $3$ is:

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