A four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?  Select the correct option:	A) ¼	B) ½	C) ¾	D) 15/16	E) 1/16
1) probability approach:   Probability of at least one even digit = Probability of one even digit + Probability of two even digits + Probability of three even digits + Probability of four even digits   I have four even digits: 0,2,6,8 and four odd digits: 3,5,7,9   So the probability of even digit is 1/21/2, as well of odd digit.   P1 = 1/21/2   P2 = 1/22=1/41/22=1/4   P3 = 1/23=1/81/23=1/8   P4 = 1/24=1/161/24=1/16   P = P1+P2+P3+P4 = 15/1615/16   2) combinatorial approach:   This approach is missing.   3) reversal probability approach:   Probability of at least one even digit = 1 - Probability of four odd digits   P = 1−1/24=15/16   ANSWER:15/16
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A four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?