Probability: Four-digit safe code

[geekybiz1]

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?

a)¼
b)½
c)¾
d)15/16
e)1/16

Again, requesting to provide explanation. Official Answer later.

[manish8109]

Ans D 15/16

P(no even digit) = 4/8 * 4/8 * 4/8 * 4/8= 1/16

P(atleast 1 even digit)= 1-1/16 = 15/16 (ANS)

[shud]

Answer is D.

total digits are = 0,(1,2,3,4), 5,6, 7,8,9
excluded in bracket and marked are allowed even
even = 3 odd = 3

for frist place of the code -prob ( no even) = 3/6 = 1/2
so no even code = 1/2*1/2*1/2*1/2 = 1/16

so at least one even = 1-1/16 = 15/16

[GMAT-HELP]

Answer is D.

total digits are = 0,(1,2,3,4), 5,6, 7,8,9
excluded in bracket and marked are allowed even
even = 3 odd = 3

for frist place of the code -prob ( no even) = 3/6 = 1/2
so no even code = 1/2*1/2*1/2*1/2 = 1/16

so at least one even = 1-1/16 = 15/16

The digits allowed are 2,3,5,6,7,8,9,0 (only 1 & 4 are not allowed)
Odd digits are 3,5,7,9
Atleast 1 even digit = 1 - all odd digits
Assuming Ist digit can take "0", the prob of all odd digits = 4/8*4/8*4/8*4/8
=> 1 - 1/16 = 15/16
Ans D.

[shud]

Thanks GMAT-HELP
what a blunder..

[geekybiz1]

Thanks frnds. D it is.
Crystal clear to me now.