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

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

A. \(\dfrac12\)

B. \(\dfrac{31}{36}\)

C. \(\dfrac{49}{54}\)

D. \(\dfrac78\)

E. \(\dfrac{11}{12}\)

Answer: B

Source: Manhattan GMAT

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The product of the digits will only be odd if all three digits are odd. If we want all of our digits to be odd, we'd have 5 choices for each digit, so there will be 5^3 = 125 such three-digit numbers.

For the remaining 900 - 125 = 775 three-digit numbers, the product of the digits will be even. So if we pick one of the 900 three-digit numbers at random, the probability it will have an even product of its digits is 775/900. The numerator and denominator are both divisible by 25, so we can cancel 25, to get 31/36.
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