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I was studying the binomial distribution and found this question below

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:

(A)$\frac{1}{2^n}$

(B)$1-\frac{1}{n}$

(C)$\frac{1}{n!}$

(D)$1-\frac{1}{2^n}$

Can I visualise this question as say we have a set of $2^n$ elements corresponding to set of all strings we can make by tossing a fair coin n times repeatedly.

Now, from this $2^n$ strings, I pick up one.Now there are remaining $2^n-1$ Strings.Now, the probability that when I pick up a string from remaining strings, won't match with the one that I picked up first will be

$\frac{2^n-1}{2^n}=1-\frac{1}{2^n}$

Is my visualisation correct?

user3767495
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Yes, you can.

The reason why you can view it as such set is due to you are tossing a fair coin independently. Hence each possible string is equally likely.

Siong Thye Goh
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  • @Siong-Okay, since each coin toss is independent and each possible string is equally likely, that's why I was able to use the formula $\frac{n(E)}{n(S)}$ where n(E) is the number of favourable outcomes and n(S) is total number of outcomes in the sample space. Right? – user3767495 Jul 08 '18 at 11:05
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    yup, your understanding is correct. – Siong Thye Goh Jul 08 '18 at 11:06