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A hacker knows that the password to the TIFR server is $10$-letter string consisting of lower-case letters from the English alphabet. He guesses a set of $5$ distinct $10$-letter strings (with lower-case letters) uniformly at random. What is the probability that one of the guesses of the hacker is correct password?

  1. $\frac{5}{(26)^{10}}$
  2. $1-\left (1- \frac{1}{(26)^{10}}\right )^{5} $
  3. $1-\left\{\left (\frac{(26)^{10}- 1}{(26)^{10}}\right )\left (\frac{(26)^{10}- 2}{(26)^{10}}\right )\left (\frac{(26)^{10}- 3}{(26)^{10}}\right )\left (\frac{(26)^{10}- 4}{(26)^{10}}\right )\left (\frac{(26)^{10}- 5}{(26)^{10}}\right )\right\}$
  4. $ \frac{1}{(26)^{10}}$
  5. None of the above
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Total number of $5$ string sets is $\binom{26^{10}}{5}$

 Out of which, total $\binom{26^{10}-1}{4}$ sets will contain password.

So required probability that his guessed set contains the password will be $\frac{\binom{26^{10}-1}{4}}{\binom{26^{10}}{5}}$.

And if you simplify $\frac{\binom{n-1}{r-1}}{\binom{n}{r}}$, it will be $\frac{r}{n}$ $\rightarrow$ $\frac{5}{26^{10}}$.

$\rightarrow$ So, Option $A$
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Total number of Passwords possible- 2610. (As each password character have 26 choices to choose from).

He guesses a set of 5 different 10 letter strings. => He guesses five different passwords.
Therefore, 

Therefore, Probability that one of the guesses of the hacker is correct password = 5 / 2610 (Option A)

Answer:

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