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16 Apr 2006, 18:59
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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?
a) ¼
b) ½
c) ¾
d) 15/16
e) 1/16
a) ¼
b) ½
c) ¾
d) 15/16
e) 1/16
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16 Apr 2006, 19:29
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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..
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..
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
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
My mistake. I read the Q as doesn't contain digits 1 to 4.
I will edit my post!
