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17 Aug 2005, 09:33
Kudos
Bookmarks
hello everyone
i found this one interesting .It comes from the web
John wrote a phone number on a note that was later lost , John can remenber that the number had 7 digits, the digit 1 appear ed in the last three places and 0 did not appear at all . what is the probability that the phone numbet contains at least two prime numbers
15/16
11/16
11/12
1/2
5/8
PLZ shows full work and reasoning
will post the OA LATER
i found this one interesting .It comes from the web
John wrote a phone number on a note that was later lost , John can remenber that the number had 7 digits, the digit 1 appear ed in the last three places and 0 did not appear at all . what is the probability that the phone numbet contains at least two prime numbers
15/16
11/16
11/12
1/2
5/8
PLZ shows full work and reasoning
will post the OA LATER
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17 Aug 2005, 10:01
Kudos
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I'm assuming the following:
1 is not repeated other than in the last 3 digits and that repeats of other numbers are allowed. Given that:
We have a number that is xxxx111 where the 4 xs are unknown and can be any of the numbers 2-9. Listing them we have:
Primes - 2,3,5,7
Non-primes - 4,6,8,9
Total number of different numbers = 8^4 since each x can be filled by any of the 8 numbers 2 thru 9
# of times we have >=2 primes = Total - (# of times we have no primes) - (# of times we have 1 prime)
# of times we have no primes = 4^4
# of times we have 1 prime = (Select the prime) * (Select the spot) * (Fill the other 3 spots with any of 4,6,8,9) = C(4,1) * C(4,1) * 4^3 = 4^5
So the required probability = 1 - (4^4 + 4^5)/(8^4) = 1 - 5/16 = 11/16
1 is not repeated other than in the last 3 digits and that repeats of other numbers are allowed. Given that:
We have a number that is xxxx111 where the 4 xs are unknown and can be any of the numbers 2-9. Listing them we have:
Primes - 2,3,5,7
Non-primes - 4,6,8,9
Total number of different numbers = 8^4 since each x can be filled by any of the 8 numbers 2 thru 9
# of times we have >=2 primes = Total - (# of times we have no primes) - (# of times we have 1 prime)
# of times we have no primes = 4^4
# of times we have 1 prime = (Select the prime) * (Select the spot) * (Fill the other 3 spots with any of 4,6,8,9) = C(4,1) * C(4,1) * 4^3 = 4^5
So the required probability = 1 - (4^4 + 4^5)/(8^4) = 1 - 5/16 = 11/16
