A three-digit positive integer is chosen at random. What is the probability that the product of its digits is even?

[BTGmoderatorDC]

A three-digit positive integer is chosen at random. What is the probability that the product of its digits is even?

A. 1/2
B. 31/36
C. 49/54
D. 7/8
E. 11/12


OA B

Source: Manhattan Prep

[swerve]

A three-digit positive integer is chosen at random. What is the probability that the product of its digits is even?

A. 1/2
B. 31/36
C. 49/54
D. 7/8
E. 11/12


OA B

Source: Manhattan Prep

A quick way to solve it consists of computing the probability of the complementary event:

P(three odd digits) \(= \dfrac{5}{9}\cdot \dfrac{5}{10}\cdot \dfrac{5}{10} = \dfrac{5}{36} \)(first digit cannot be \(0,\) so we have a total of \(9\) digits, \(4\) even and \(5\) odd).

Therefore, P(at least one even digit) \(= 1 - \dfrac{5}{36} = \dfrac{31}{36}.\)

Hence, B