Denise is trying to open a safe whose combination she does

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: \(\frac{3999}{4000}*\frac{3998}{3999}*...*\frac{3925}{3926}\) every denominator but the first will cancel out and every nominator but the last will cancel out as well.

We'll get \(\frac{3925}{4000}=\frac{157}{160}\).

Answer: C.
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If Denise does not pick the correct combination in first 75 try, then the correct combination is one of remaining 3925. So Probability = 3925/4000 = 157/160.

answer is C.

OMG this looks tough, sorry I dont have a solution.

isn't it

\((3999/4000)^{75}\)

???

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

That is how you do it.

75/4000 is the probability that she gets is. That means that 1-(75/4000) is the probability that she doesn't get it. 1-(75/4000)=3925/4000 which simplifies into 157/160.

If you struggle to simplify (like me), start by dividing 4000 by 160 since you have a 3/5th chance of having 160 as the denominator in the answer. 4000/160=25. Now divide 3925 by 25 and you get 157. That means the answer is 157/160.

Hope my simple minded way of explaining helps!!

Hmm interesting- i went with the 1-75/4000 approach but bunuel seems to think out of the box!

1 - (75/4000) (1 minus probability of picking correct combination )
3925/4000
157/160

Hence C

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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probablity of not getting the right combination in 75 attempts = 1 - prob of getting right combination in all 75 attempts
= 1 - [ 1/4000 + 1/4000 +..75times....1/4000]

1 - ( 75/4000)

= 157/160

probability of one of the trials being the right one. 75c1/4000. so of not being, 1-75c1/4000 = 157/160

Hi guys - just a short comment:

3925 / 4000 = (5)*(5)*(157) / (2)*(2)*(2)*(5)*(5)*(5)*(4)

Cross out 2*(5) --> calculate

.... or see that C is the only option containing 157 in the numerator.

we get 3925/4000,

so it can be B or C. We need 4000:160 to know divisor for 3925, it is 25

3925/25=157

C

Solution

Given:

Denise is trying to open a safe whose combination she does not know
Safe has 4000 possible combinations.
Denise can try 75 different possibilities.

Approach and Working:

P( Denise does not pick the correct combination)= P(1st attempt incorrect combination) * P(2nd attempt incorrect)*………….* P(75th attempt incorrect)
P( Denise does not pick the correct combination)= 3999/4000*3998/3999*……..*3925/3926 = 3925/4000 = 157/160

Therefore, the correct answer is option C.
Answer: C
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Probability of not picking correctly = 1 - Probability of picking correctly
Probability of picking correctly:
Getting it right in first choice: 1/4000
Getting it right in the second choice(wrong in first attempt, right in the second): 3999/4000 x 1/3999 = 1/4000
Getting it right in the third choice(wrong in first attempt, wrong in the second, right in the third): 3999/4000 x 3998/3999 x 1/3998 = 1/4000
...
Getting it right in the 75th choice = 1/4000

Total probability of getting it right = Right in first choice OR right in the second...OR right in the 75th = 1/4000+ 1/4000.. 75 times = 75/4000
Therefore, Probability of not picking correctly = 1 - 75/4000 = 157/160

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