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Intern
Joined: 01 Aug 2003
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01 Aug 2003, 08:47
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Pls help
A certain telephone number has 7 digits.If the
telephone number has the digit 0 exactly three times
and number one is not used at all what is the
probability that phone number contains 1 or more prime
numbers?

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Manager
Joined: 02 Jul 2003
Posts: 58

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01 Aug 2003, 09:56
eliminate three digits because 3 are zero. so last four digits could be 2,3.,4,5,6,7,8,9 for a totatl of 8^4 choices. Prime numbers are 2,3,5,7 so favorable outcomes are 4^4. so answer (hopefully)is 4^4/8^4.

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SVP
Joined: 03 Feb 2003
Posts: 1603

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01 Aug 2003, 12:08
My approach:

P(at least 1 prime)=1-P(no primes)

Total outcomes: 8*8*8*8*1*1*1 - but this case is when 3 zeroes are the last digits. But it is not given. We have multiply by 6C3=20. So, 20*8^4

Favorable outcomes: 4*4*4*4*1*1*1*6C3=20*4^4

P=1-(4^4/8^4)=1-(1/16)=15/16

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Manager
Joined: 03 Jun 2003
Posts: 83

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Location: Uruguay

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01 Aug 2003, 12:54
My try,

P(at least one prime) = 1- P (no prime)

7 numbers of which 3 are 0┬┤s and no 1┬┤s

So we have 4 numbers left to find out

primes (2,3,5,7) not primes (4,6,8,9)

first number P (not prime) = 4/8 = 1/2

so 1- (1/2)^4 = 15/16

Kudos [?]: 1 [0], given: 0

01 Aug 2003, 12:54
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