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A brief-case lock consists of 3 rolling disks, with the

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A brief-case lock consists of 3 rolling disks, with the [#permalink] New post 05 Jul 2004, 00:07
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A brief-case lock consists of 3 rolling disks, with the first one being marked 0—9, the second 0—6, the third 1—5. What is the probability of not opening the lock for the first try?
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 [#permalink] New post 05 Jul 2004, 00:37
A brief-case lock consists of 3 rolling disks, with the first one being marked 0—9, the second 0—6, the third 1—5. What is the probability of not opening the lock for the first try?


the first roller can have 10 possibilites =10
the second can have 7 possibilities = 7
the third can have 5 possibilities =5

Total possiblities = (10x7x5)=350

Probability is 1- (1/total)=1-(1/350) = 349/350

Is it correct.

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 [#permalink] New post 05 Jul 2004, 09:22
I agree with your solution Carsen. But (note the use of But :-) ), as it is Stoylar who has posted the q., there could be something fishy abt. it!!
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 [#permalink] New post 05 Jul 2004, 09:30
I am not quite sure, but I think the answer should be this way:

probability of not getting the first number right * probability of not getting the second number right * probability of not getting the third number right= p(1-(1/10))*p(1-(1/7))*p(1-(1/5)

=> 9/10*6/7*4/5=216/350

Correct me If I am mistaken.
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 [#permalink] New post 05 Jul 2004, 10:07
dr_sabr:

the problem in your approach is that your approach assumes that all the three numbers should not be correct, instead of the combination.

say, the correct comb. is 014

it is possible that 0 is in the first place but still the combination is incorrect (like 025).

I suggest you consider the event definition in this case is to "open the lock" for this the total conbimations possible is 10*7*5 out of these there is one combination opens the lock. Therefore the number of reqd combinations is 10*7*5 - 1 and hence the probability is
(10*7*5 -1)/10*7*5
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 [#permalink] New post 05 Jul 2004, 12:19
Vithal wrote:
dr_sabr:

the problem in your approach is that your approach assumes that all the three numbers should not be correct, instead of the combination.

say, the correct comb. is 014

it is possible that 0 is in the first place but still the combination is incorrect (like 025).

I suggest you consider the event definition in this case is to "open the lock" for this the total conbimations possible is 10*7*5 out of these there is one combination opens the lock. Therefore the number of reqd combinations is 10*7*5 - 1 and hence the probability is
(10*7*5 -1)/10*7*5


Thank you Vithal, that helps :wink:
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  [#permalink] 05 Jul 2004, 12:19
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