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

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