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Probability

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by f2001290 » Tue May 29, 2007 7:30 am
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?

I don't have an OA.

P(atleast one digit is even) = 1 - P(all digit odds)

We have 4 odd digits excluding 1 (3,5,7,9).
So, possible 4 digit odd numbers = 4*4*4*4
All possible 4 digit numbers = 7*8*8*8 (Ensure that 1st digit is not 0 and exclude 1 and 4)

P(all digits odd) = 4*4*4*4/7*8*8*8 = 1/14
P(atleast one digit is even) = 1 - 1/14 = 13/14

Any different solution for this . I am very bad at probability
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by RAGS » Tue May 29, 2007 11:03 am
4 digit safe code without 1 and 4
possible digits for each digit 0,2,3,5,6,7,8,9 ie 8 ways
total 8*8*8*8

consider all odd digits so possible digits 3,5,7,9 ie 4 ways
total 4*4*4*4

So atleast one even is= 8*8*8*8 -4*4*4*4 / 8*8*8*8

= 1 - 1/16 = 15/16

Why should the first digit not be zero????????????
If It Is To Be It Is Up To Me
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by f2001290 » Tue May 29, 2007 10:18 pm
If 1st digit is zero, then it becomes a 3 digit number.

0123 is a 3 digit number.

Rags, I may be wrong (most of the time) :wink: . The most unfortunate thing is that I don't have an OA
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by gabriel » Wed May 30, 2007 8:46 am
... i agree with RAGS ... 0 can be the first number .. wen u say that 0 as the first number wuld reduce it to a 3 digit number .. that is only true wen u consider the numerical value of the 4 digit number ... over here the numerical value is not important but the combination of different numbers is.. so 0 can be the first number ... btw my bag which had a number lock used to open with a 0000 :D
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