A threedigit 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
A threedigit positive integer is chosen at random. What is the probability that the product of its digits is even?
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 Ian Stewart
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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 threedigit numbers.
For the remaining 900  125 = 775 threedigit numbers, the product of the digits will be even. So if we pick one of the 900 threedigit 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.
For the remaining 900  125 = 775 threedigit numbers, the product of the digits will be even. So if we pick one of the 900 threedigit 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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