# Thread: Probability: Four-digit safe code

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## Probability: Four-digit safe code

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.

2. Good post? |
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)

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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

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Originally Posted by shud

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.

5. Good post? |
Thanks GMAT-HELP
what a blunder..

6. Good post? |
Thanks frnds. D it is.
Crystal clear to me now.

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