To solve this problem, we need to determine if both statements about probabilities in forming words from the letters of "SYLLABUS" are correct.
Calculate the total number of distinct 8-letter words formed from "SYLLABUS".
Letters in "SYLLABUS": S, Y, L, L, A, B, U, S.
Since 'L' and 'S' are repeated, the total number of arrangements is given by:
\(\frac{8!}{2! \times 2!}\)
Calculating this:
\(\frac{40320}{4} = 10080\)
Check Statement I: "The probability that the word contains the two S's together is \(\frac{1}{4}\)."
Treat the two 'S's as a single unit. Thus, we have 7 units to arrange: 'SS', Y, L, L, A, B, U.
The number of ways to arrange these units is:
\(\frac{7!}{2!}\)
Calculating this:
\(\frac{5040}{2} = 2520\)
The probability is then:
\(\frac{2520}{10080} = \frac{1}{4}\)
Thus, Statement I is correct.
Check Statement II: "The probability that the word begins and ends with L is \(\frac{1}{28}\)."
Fix 'L' at both ends: L _ _ _ _ _ _ L. Now, we arrange the remaining: S, S, Y, A, B, U (6 unique letters).
The number of ways to arrange these letters is:
\(\frac{6!}{2!}\)
Calculating this:
\(\frac{720}{2} = 360\)
The probability is then:
\(\frac{360}{10080} = \frac{1}{28}\)
Thus, Statement II is correct.
Conclusion: Both Statement I and Statement II are true, so the correct answer is "Both Statement I and Statement II are true".

