ISRO 2009-ECE Probability

Question

A man with n keys wants to open a lock. He tries his keys at random.

The expected number of attempts for his success is (keys are replaced after every attempt)

a) n/2

b) n

c) √ n

d) None of the above

Comments

I took small example of four key and my approach is

1(1/4) + 2( 3/4 ×1/3)+3(3/4 × 2/3 × 1/2) + 4 (3/4 × 2/3 ×1/2 × 1)

= 1/4 ( 1+2+3+4)= 4×5/(4×2) = 5/2

So I think answer should be (n+1)/2 and so none of these.

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what i think is -

for one attempt ,probability is 1/n

for second attempt,probablity is (1-1/n) * 1/n

for third attempt prob is (1-1/n)² * 1/n

and so on..

so,expectation is E[x] =  1* 1/n + 2 ( (1-1/n) * 1/n) + 3* (1-1/n)² * 1/n +....

multiplying by (1-1/n) = 1/n [  (1-1/n) + 2 * (1-1/n)² + 3* (1-1/n)³ + ...]

subtracting above two we get ,E[x] - (1-1/n) E[x] = 1/n * (1/1-(1-1/n))

1/n E[x] =1

so,E[X] =n

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Expectation can be thought as mean or average. So supoose you have n keys then I dont think on average we need n trials. On average answer will be (n+1)/2. Also try 4 keys example and accordingly I think n+1/2 should be answer.

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@rahul,it is given that keys are replaced after every attempt,so why are u taking probabilitie as relative.

as you are showing that for second attempt,first attempt fauilure is 3/4 but how success for second attempt is 1/3??as after fauilure of first key,it will be replaced in the lot.,so it should be 1/4 only

correct me if i am wrong pls

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Also keys are replaced affter attempts. So (1-1/n) and then 1/n is inappropriate

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Lets say i have A,B,C,D and C is my key to open lock. Now for succes on two attempts I can select A,B,D hence 3/4 after that attempt I have only 3 keys remaining and for succes on two attempts in this attempt I have to select C which gives 1/3.

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as keys are always replaced ,so sample space will always be n..is'nt it so??

if it would have been without replacement then only,sample space starts decreasing by 1

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Okay I got what you intend to say. But I think "replace" should not be used here otherwise it is possible that we go to infinte series and we may never get success.

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for your A.B.C.D example...

in first attempt,you can choose btw A,B AND D,hence 3/4.suppose ,you choose A.it is given in question that key is replaced,hence again,A will be added to the set of keys and now,from A,B,C,D ,we will choose C,so prob of success at second attempt is 1/4

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yes..i am only gng by question's statement

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Okay actually i have misinterpreted the question. You are correct. B is the correct answer I also verified it and question is also correct.

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We can also use arithmetico geometric series formula for easy evaluation.

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thanks for confirming :)

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i dun know any such formula..nt good at these formulas..again if u have a good link then pls share  :P

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It is good if you can solve without formula so that exam time you don't need to mug them up.☺ This formula is quite handy

https://en.m.wikipedia.org/wiki/Arithmetico-geometric_sequence

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Answer is n only

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by arithmetico geometric series answer is coming n and i think we can directly use geometric distribution with probablity p=1/n and hence number of attempts= 1/p=n.

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@sushmita how to identify it there is geometric distibution???

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here we are asked number of trials and whenever no of trails are asked it is geometric distribution.

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Thanks @sushmita. 

https://en.wikipedia.org/wiki/Geometric_distribution?wprov=sfla1

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@sushmita,is it true in every case??