Binomial Distribution question based on n bit string

Question

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?

Answer 1

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.