- f2001290
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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?
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
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












