rocketscience
Hello,
I got this question from Thursdays with Ron. I wonder if someone could explain how I get the solution without the "1- not desired outcome" trick? What do you think is the problem's difficulty level?
Thanks!
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
Any a three-digit positive integer has AT LEAST one even number. In this sense, any number that doesn't have an even digit is odd.
Therea are 5 odds digits: 1,3, 5, 7, and 9.
So, the total number of numbers whose product is odd (all digits are odd) is: 5*5*5 =125
Also, the total number of numbers with three digits is: 9*10*10 = 900. (The hundreds digits only has 9 possibilities. It cannoy be 0).
Ok, we have everything, let's calculate:
\(\frac{(9*10*10 - 5*5*5)}{(9*10*10)} = 1 - \frac{(5*5*5)}{(9*10*10)} = 1 - \frac{5}{36} = \frac{31}{36}\)
Answer B.