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A Four digit safe code does not contain the digits 1 and 4

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A Four digit safe code does not contain the digits 1 and 4 [#permalink] New post 16 Apr 2006, 17:59
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
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 [#permalink] New post 16 Apr 2006, 18:29
i am getting 13/14.

1 -P(all odd).

Available numbers: {0,2,3,5,6,7,8,9}
1-{4/7*4/8*4/8*4/8}

4/7, because the first digit cant be 0, so 4 out 7 numbers, and the rest are 4/8..

But I am not getting the required..
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 [#permalink] New post 16 Apr 2006, 20:06
P(atleast 1 even) = 1 - P(all odd)

Choice of digits available = {0,2,3,5,6,7,8,9} --> Total 8
Odd digits = {3,5,7,9}
P(all odd) = 4/8*4/8*4/8*4/8 = 1/16

Reqd Prob = 1 - 1/16 = 15/16
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Last edited by giddi77 on 16 Apr 2006, 20:16, edited 1 time in total.
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 [#permalink] New post 16 Apr 2006, 20:08
how about 2 and 3? You said 1 and 4.. Did you mean 1 through 4 not included?

Thanks..
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 [#permalink] New post 16 Apr 2006, 20:15
willget800 wrote:
how about 2 and 3? You said 1 and 4.. Did you mean 1 through 4 not included?

Thanks..


Oops! :beat :wall My mistake. I read the Q as doesn't contain digits 1 to 4.

But My answer is correct though :lol: I will edit my post!
Availabe digits = {0,2,3,5,6,7,8,9} --> 8
Odd = {3,5,7,9} --> 4
P(odd) = 4/8*4/8*4/8*4/8 = 1/16
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 [#permalink] New post 16 Apr 2006, 20:18
willget800 wrote:
how about 2 and 3? You said 1 and 4.. Did you mean 1 through 4 not included?

Thanks..


Yeah, I think 2 and 3 should also be included. The answer remains the same though.

total of 8 numbers to choose from. (0-9 except 1&4)

4 even and 4 odd.

p = 1 - 4^4 / 8^4

= 15/16

D.

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 [#permalink] New post 17 Apr 2006, 10:08
p(NOT HAVING SINGLE EVEN DIGIT) = 4/8*4/8*4/8*4/8 = 1/16

p(HAVING ATLEAST ONE SINGLE EVEN DIGIT) = 1-1/16 = 15/16
  [#permalink] 17 Apr 2006, 10:08
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A Four digit safe code does not contain the digits 1 and 4

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