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)/4000Alternative 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)
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry
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