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# PS

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Intern
Joined: 01 Aug 2003
Posts: 48
Location: Mumbai
Followers: 1

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

PS [#permalink]  01 Aug 2003, 07:47
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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?
Manager
Joined: 02 Jul 2003
Posts: 58
Followers: 1

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

I'll take a shot [#permalink]  01 Aug 2003, 08:56
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.
SVP
Joined: 03 Feb 2003
Posts: 1607
Followers: 6

Kudos [?]: 84 [0], given: 0

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
Manager
Joined: 03 Jun 2003
Posts: 84
Location: Uruguay
Followers: 1

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

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
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PS 1 15 Dec 2005, 14:59
PS 4 21 Oct 2005, 09:22
ps 2 11 Oct 2005, 15:04
ps 3 11 Oct 2005, 14:58
Ps 4 17 Sep 2005, 04:31
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