Q)A security consultant is recommending that one of his clients install a keypad with a certain no. of distinct characters at the front and back door of his home.the keypad is designed so that each character is pressed once in a certain order.if the consultant advises that the probability of someone guessing the code in one attempt to be less than 0.01%.what is the minimum no. of characters the keypad needs?
6
7
8
9
10
kindly explain the solution.I am not able to understand it.
For 6 Characters:
Total distinct combinations can be:
6x5x4x3x2x1 = 720 possible combinations.
To guess on the 1st attempt the probability will be
1/720 ie 0.00138888…
To percent: 0.001388… x 100 = 0.13888%
Which is over 0.01%
So next:
For 7
Total: 720 x 7 = 5040
P(X) = 1/5040 = 0.000198
To percent: 0.000198 x 100 = 0.0198%
Still greater than 0.01%
So next
For 8
Total = 5040 x 8 = 40320
P(X) = 1/40320 = 0.0000248
To percent: 0.0000248 * 100 = 0.00248%
Which is less than 0.01
Hence the Correct answer should be 8
Alternatively:
You could go the other way around:
To get a probability of 0.01% on 1st attempt
ie: Probality = 1/(Total distinct combinations) < 0.01%
ie: Total no of combinations > 1/0.01% = 1/0.0001 = 10,000
So if the total no of combinations exceed 10,000 we know the probability of getting 1 attempt correct is less than 0.01%
So for 6 characters:
total combinations = 6x5x4x3x2x1 = 720 <10,000(naah)
For 7:
total combinations = 7! = 5040 < 10000 (naah)
For 8
total combinations = 8! = 40,320 > 10000 (yess)
Hence 8
Thanks a lot!