If you were to dial any 7 digits on a telephone in random order,
what is the probability that you will dial your own phone number?
Assume that your telephone number is of 7-digits.
March 2, 2010
Phone number permutation
If you were to dial any 7 digits on a telephone in random order,
what is the probability that you will dial your own phone number?
Assume that your telephone number is of 7-digits.
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10 comments:
1 in (10^7)
1 in 9x10^6
1 in 9000000 cuz its observation over the total frequency and your phone number is one number and the total possible numbers is 900 0000 (100 0000 to 999 9999)
1/(10^7), since there are 10 numbers (counting 0), and there are 7 digits in the number.
It is 1 in 9x10^6 (1 in 9,000,000) because no-one's local phone number begins with zero. Zeroes are reserved for signalling the exchange that you are making a special type of call; eg emergency (000), out-of-area (0n nnn nnnn), mobile/cell (0nnn nnn nnn), etc.
1/(7x10^N).1 represents your phone number,7 represents the number of digits of your phone number, 10 represents the numbers that can be dial in a telephone and the N represents that how many times do you dial before you could possibly dial your phone number.....
To start off , let us tackle numbers with 2 digit only.
1st digit 2nd digit
can be 1-9 can be 0-9
because nos don't start with 0 for that will make the number a one digit entity.So toally 9*10 = 90 nos
for 3 digit nos
1-9 0-9 0-9
9*10*10 = 900
Similarly for 7 digit no,it will be
9*10*10*10*10*10*10*10
= 9 * 10^6
= 9* 10 ^ (no of digits -1 )
Hence our probability is
1 / 9*10^(7-1) = 1/ 9*10^6
since 1st digit can not be zero
total no of ways=9*10*10*10*10*10*10
=9*10^6
probability=1/9*10^6
one mill of a chance
10^7
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