The answer is D. You can very quickly narrow the answers to C or D by using simple logic. Because an equal number of odd and even digits are available for selection, it's very likely that at some point an even digit will be picked. The answer must be greater than 50%, close to 1. Anyway, let's solve.
Whenever you are asked to find the probability of "at least one", find the probability of never and subtract it from 1. P(at least one) + p(never)= 1
Find the probability that there is no even digit (only odd). Probability is =(# of ways to get what you want)/(total # of ways)
Since digits can be repeated, but the digit 1 cannot be used, each of the 4 digits can be (3, 5, 7 or 9). Because there are 4 digits, the number of codes with only odd digits is 4*4*4*4.
Find the total number of codes possible. Each digit can be anything but a 1 or 4, so there are 8 options total. Because there are 4 digits in the code, the total number of codes possible is 8*8*8*8.
The probability that there is no even digit (only odd) is (4*4*4*4)/(8*8*8*8) = (1/2)^4 = 1/16.
Remember that P(at least one) + p(never)= 1. Since the probablity that we never pick an even is 1/16, the probabilty that we pick at least one even is 15/16.
If you have trouble with similar questions, set topics='combinatorics' and difficulty='700+' in the Drill Engine. Such similar questions include
GMATPrep Question 1233, and
GMATPrep Question 1474
Good Luck,
-Patrick
