Three numbers are chosen at random from 1 to 20 , then the probability that the sum of three numbers is…
Three numbers are chosen at random from 1 to 20 , then the probability that the sum of three numbers is divisible by 3 is
$\frac{1}{114}$
$\frac{147}{342}$
$\frac{16}{47}$
$\frac{32}{85}$
Solution
Total number ways we can choose three integers from 20 integers $={ }^{20} C_3$
(I) We will get all three integers are multiple of three $(3,6,9,12,15,18)={ }^6 C_3$
(II) We will get sum divisible by 3, if all three numbers when divided by 3 givens remainder of $1(1,4,7,10$, $13,16,19)={ }^7 C_3$
(III) We will get sum divisible by , if all three numbers are when divided by 3, gives a remainder of $2(2,5$, $8,11,14,17,20)={ }^7 C_3$
(IV) We will get sum divisible by 3, if one of themis divisible by 3, one of them when divided by 3 gives a remainder of 1 , one of them when divided by 3, gives a remainder $2={ }^7 C_1 \times{ }^7 C_1 \times{ }^6 C_1$
Probability
$=\frac{{ }^6 C_3+{ }^7 C_3+{ }^7 C_3+{ }^7 C_1 \times{ }^7 C_1 \times{ }^6 C_1}{{ }^{20} C_3}=\frac{32}{85}$