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# A certain telephone number has 7 digits. If the telephone

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CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

A certain telephone number has 7 digits. If the telephone [#permalink]  01 Oct 2003, 16:47
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A certain telephone number has 7 digits. If the telephone number
has the digit 0 exactly three times, and the number one is not used at all,
what is the probability that the phone number contains 1 or more prime
digits?
SVP
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

P=1–P(no primes)

the total number of combinations: _ _ _ _ _ _ _
3 zeroes=7C3, but this includes a zero on the first place. Eliminate this fact: 7C3–6C3. Four other positions can be filled with 8 digits each. In total, we have 8*8*8*8*[7C3–6C3]

the number of combinaions having no primes 4*4*4*4*[7C3–6C3]

P=1–[4*4*4*4]/[8*8*8*8]=1–1/16=15/16
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

The first digit cannot be zero and since 1 is not used the first digit can range from 2-9 = 8 combinations
In the remaining 6 places we have 3 zeros in 3! * 4! ways. The other three places can take any values from 2 to 9 ( 8 values per position)
Total combinations are 8 * 3! * 4! * 8^3 = 3!*4!*8^4
Desired combinations are atleast one prime = 1-P(no prime)
The only non primes are 2,4,6,8
Using the same technique com binations with no primes = 4*3!*4!*4^3
= 3!*4!*4^4

p = 1- ( 3! * 4! * 4^4 ) / ( 3! * 4! * 8^4 ) = 15/16
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