Denise is trying to open a safe whose combination she does not know. IF the safe has 4000 possible combinations,and she can try 75 different possibilities,what is the probability that she does not pick the one correct combination.

A. 1
B. 159/160
C. 157/160
D. 3/160
E. 0

When trying the first time the probability Denise doesn't pick the correct combination=3999/4000
Second time, as the total number of possible combinations reduced by one, not picking the right one would be 3998/3999.
Third time 3997/3998
...
And the same 75 times.

So we get: 3999/4000*3998/3999*...*3925/3926 every denominator but the first will cancel out and every nominator but the last will cancel out as well.

We'll get 3925/4000=157/160.

Answer: C.
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There is a huge probability that i may be embarrassing my self on this one .

But anyways .
Why is it not 1-75/4000 that simple, i know the answer is coming up the same

probability of getting the right choice is 75/4000 so not getting it right is 1-(75/4000)

Please help i'm really bad at probability
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Bunuel, i did the same way as utkarshlavania. is there anything wrong with that?

thanks.
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Hello all, can we not use Combination to solve this problem? cos problem doesnt say first 75. So why cant we select 75 out of 4000 by combination?I know the probability solution is much easy, but just to clarify my understanding of Combinations.
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I have a query here :

1 - (75/4000) (1 minus probability of picking correct combination )

Why isn't 75/4000 the probability of picking incorrect combinations, because all the 75 combinations can be wrong too, i.e. how can we assume that out of 75, one could be correct combination ?
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Think of it more simply. Lets start by taking the case where you have only 1 guess :

Probability of getting it right = 1/4000
Probability of getting it wrong = 3999/4000

Now take the case you have 2 guesses :

Probability of getting it right = Probability u get it in first go + Probability u get it in second go
= (1/4000) + (3999/4000)*(1/3999)
= (1/4000) + (1/4000) = (2/4000)
Probability of getting it wrong = 1 - Probability of getting it right = 1 - (2/4000) = 3998/4000

If you do this again, you will see probability of getting it right in k turns = (k/4000) and not getting it right is (4000-k)/4000

Alternative Approach

Imagine you right down all 4000 combinations of the safe one after the other.
And also that you right it down in every possible order.
Now for every order you have written down, the kth number is the kth try you will be making.
The rhetorical question I ask is how many times is the correct number appearing in exactly the kth position, in all your orderings ?
The trick here is to grasp the fact that the number of times you get the right combination in the kth slot is independent of k, by symmetry.
The right number is equally likely to be in the first slot as it is in the second as it is in the third and so on so forth.
And since for any ordering, the kth number is nothing but the kth try, and the chances that the kth number is the correct number are equal for all values of k. We can conclude that the probability that you get the number correct in the kth try is exactly (1/4000).

Hence getting it right in 75 tries = Getting in right in the first try + Getting it right in the second try + ... + Getting it right in the 75th try = (1/4000) + (1/4000) + ... + (1/4000) = (75/4000)
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I got it "C" by following procedure,Please let me know if I am wrong in taking this approach,
75 combinations are correct out of 4000 possible combination thus probability of him getting the correct combination is = 75/4000 = 3/160

So the probability of him getting the incorrect combination is 1 - (3/157) = 157/160


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