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25 votes

A random bit string of length n is constructed by tossing a fair coin n times and setting a bit to 0 or 1 depending on outcomes head and tail, respectively. The probability that two such randomly generated strings are not identical is:

  1. $\frac{1}{2^n}$
  2. $1 - \frac{1}{n}$
  3. $\frac{1}{n!}$
  4. $1 - \frac{1}{2^n}$
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Total possible selections ((2n)C1 ) * ((2n)C1 ) . Favorable outcomes   (2n) * (2n-1)

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9 Answers

50 votes
50 votes
Best answer

Answer - $D$

Suppose there are $k$ places within $n$ bit string where mismatch has occoured

Probability of this occurring is 

$^{n}C_{k}.\text{(prob. of mismatch)}^{k}\text{(prob. of match)}^{(n-k)}$

$ = ^{n}C_{k}\left(\dfrac{1}{2}\right)^{k}\left(\dfrac{1}{2}\right)^{(n-k)} $

$= ^{n}C_{k}\left(\dfrac{1}{2}\right)^{n}.$

$k$ can range from $1$ to $n$, hence the required probability $\sum \left(^{n}C_{k}\left(\frac{1}{2}\right)^{n}\right)$ where $k$ ranges from $1$ to $n$

is $\left(\dfrac{1}{2^{n}}\right)\left(2^{n} - 1\right).$


Alternatively

Probability of matching at given place $\dfrac{1}{2}.$

there are n places hence probability of matching $\dfrac{1}{2^{n}}.$

hence probability of mismatch $1 - \dfrac{1}{2^{n}}.$

edited by

3 Comments

Nice explanation by using Binomial distribution.Thanks :)
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@Arjun sir can we do like this, n places so total 2^n different strings are possible, now we have generated one string, so for identical we have only one favorable case (string must be same that we have already generated)

probability for identical 1/2^n

probability for not identical 1-1/2^n

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Expanding on the second part of the answer a bit:

Let us say we generate a random $n$ bit string as the first string. There's no restriction on this string. For the second string to match with this, each bit should match and each bit matches with a probability of $0.5$ and since they are independent, the probability of matching will be $0.5^n$. Subtract this from 1, we get probability of not matching.
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82 votes
82 votes
Total combinations of string that can be generated are 2^n. We will get one such string in the first experiment. So favourable cases for the second string are 2^n-1, so that it doesnt match with the previous generated string.

Hence Probablity= (2^n-1)/2^n= 1-1/2^n.

3 Comments

nice explanation ....
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This is very easy to understand than the selected answer. Thanks !
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can you please explain the numerator part .. i didn’t get why we are subracting -1
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14 votes
14 votes
Let the 1st string generated be 010101 and the 2nd string be 110011. (N=6)

To check whether they are totally identical, we can apply XOR to them. If the output comes as 000000 then we can be sure that the pattern is same else different.

Here we will see that xor comes as 011001 which is not equal to 000000 hence it is not same.

Now, probability of getting '0' in each of the n places is 1/2. Hence probability of getting 0 in all n places is (1/2)^n.

All other cases which contain mixture of 0's and 1's imply that the two strings are not fully identical ( they may be same at some bit position but not all).

So probability of getting different strings=1-probability of getting identical strings

=1-(1/2)^n.

Option D.

4 Comments

@MiNiPanda only God or you know what logic have you applied to solve this question.

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edited by

@chauhansunil20th

And probably the upvoters of this answer :P

P.S. You can check this one https://gateoverflow.in/1072/gate2004-78?show=159282#a159282

Similar approach to kind of similar question.

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hahahaha :D :D @MiNiPanda well said :P

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@MiNiPanda Sample space will be 2^n which itself gives all different strings. The how come we find favorable cases in this example?

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3 votes
3 votes

Two such randomly generated strings can be non identical in many ways

We have to sum them up.

As we can have 1,2,3,…. upto n many combinations

 

 

OPTION D

 

Answer:

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