Three numbers are chosen at random from the numbers $1, 2, ... 20.$ The probability that the arithmetic mean of these numbers is $4$ is

Total number of ways of choosing three numbers out of $20$ is ${ }^{20}{C}_3=1140$

n(S)=1140

E= arithmetic mean of the 3 numbers selected is 4

Let these numbers be $x_1,x_2,x_3.$

We have $\frac{1}{3}\left(x_1+x_2+x_3\right)=4,$

we get $x_1+x_2+x_3=12.$

The possible choices of $\left(x_1,x_2,x_3\right)$ and $(1,2,9),(1,3,8),(1,4,7),(1,5,6)(2,3,7),(2,4,6),(3,4,5)$

$\therefore$   probability of the required event $=\frac{7}{1140}.$