Question

A number of distinct 8-letter words are possible using the letters of the word SYLLABUS. If a word is chosen at random, then Statement I: The probability that the word contains the two S's together is $\frac{1}{4}$. Statement II: The probability that the word begins and ends with L is $\frac{1}{28}$. In the light of the above statements, choose the correct answer from the options given below:

Accepted Answer

A. Letters: S, Y, L, L, A, B, U, S (8 letters: 2 S's, 2 L's). Total words = $\frac{8!}{2!2!}$. Statement I: Treat (SS) as 1 unit. Words = $\frac{7!}{2!} \Rightarrow 	ext{Prob} = \frac{7! / 2!}{8! / (2!2!)} = \frac{2}{8} = \frac{1}{4}$. (True) Statement II: Fix L at start and end. Remaining 6 letters (S, Y, A, B, U, S with 2 S's) arranged in $\frac{6!}{2!}$ ways. $	ext{Prob} = \frac{6! / 2!}{8! / (2!2!)} = \frac{2}{8 	imes 7} = \frac{1}{28}$. (True) Both statements are true.

Probability Statements on Letter Arrangements of SYLLABUS

Explanation

Practise this under exam conditions