A three-digit number was chosen at random. What is the the probability that it is divisible by both 2 and 3.

Question

A three-digit number was chosen at random. What is the the probability that it is divisible by both 2 and 3. (A) (1/12) (B) (1/6) (C) (1/9) (D) (1/8)

Tardigrade Admin · Accepted Answer

Let the number be x, 100 ≤ x ≤ 999. (There are 900 three digits numbers) A number divisible by both 2 and 3 must be divisible by the LCM of (2,3)=6. Least value of x divisible by 6=102=6(17) Greatest value of x divisible by 6=996= 6(166) There are 150 values of x divisible by 6 . Required probability =(150/900)=(1/6)

Q. A three-digit number was chosen at random. What is the the probability that it is divisible by both 2 and 3.

$\frac{1}{12}$

$\frac{1}{6}$

$\frac{1}{9}$

$\frac{1}{8}$

Q. A three-digit number was chosen at random. What is the the probability that it is divisible by both 2 and 3.

121​

61​

91​

81​

$\frac{1}{12}$

$\frac{1}{6}$

$\frac{1}{9}$

$\frac{1}{8}$