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# The safe lock consists of three disks: the first one marked

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SVP
Joined: 03 Feb 2003
Posts: 1607
Followers: 7

Kudos [?]: 87 [0], given: 0

The safe lock consists of three disks: the first one marked [#permalink]  19 Dec 2003, 01:24
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
The safe lock consists of three disks: the first one marked from 0 to 9, the second marked with even digits, and the third marked with 1 and 0. A certain guy has a one attempt to open the safe by rotating disks randomly. Assuming that there is the only correct combination to open the safe, what is the probability that the guy will not open the safe?
Director
Joined: 13 Nov 2003
Posts: 790
Location: BULGARIA
Followers: 1

Kudos [?]: 30 [0], given: 0

anvar, agree with the method offered, but think that on second ring we should coun also 0 which is even, as far as i know, which makes the possible combinations 100( 10x5x2), not 80...
Intern
Joined: 27 Nov 2003
Posts: 34
Location: Moscow
Followers: 0

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

BG wrote:
anvar, agree with the method offered, but think that on second ring we should coun also 0 which is even, as far as i know, which makes the possible combinations 100( 10x5x2), not 80...

Thanks BG, good to know that "zero" is even
_________________

me

Intern
Joined: 22 Nov 2003
Posts: 41
Location: Serbia
Followers: 0

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

(analogy with 750+ on GMAT)
SVP
Joined: 03 Feb 2003
Posts: 1607
Followers: 7

Kudos [?]: 87 [0], given: 0

99% is OK
to be perfect, let's transform the number into 0.99.
according to math definition of probability, 0<=P<=1
SVP
Joined: 30 Oct 2003
Posts: 1794
Location: NewJersey USA
Followers: 5

Kudos [?]: 43 [0], given: 0

Probability that he will open the lock
= 1/10 * 1/5 * 1/2

probability that he wont open the lock = 1 - 1/100 = 0.99
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