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Nov 2020 p53 q5
2687
The 8 letters in the word RESERVED are arranged in a random order.
(a) Find the probability that the arrangement has V as the first letter and E as the last letter.
(b) Find the probability that the arrangement has both Rs together given that all three Es are together.
Solution
(a) The total number of ways to arrange the letters in RESERVED is \(\frac{8!}{3!2!} = 3360\) because there are 3 Es and 2 Rs.
To have V as the first letter and E as the last letter, we arrange the remaining 6 letters: R, E, S, E, R, D. The number of ways to arrange these is \(\frac{6!}{2!2!} = 180\).
The probability is \(\frac{180}{3360} = \frac{3}{56}\) or 0.0536.
(b) Consider Rs together and Es together as single units: [RR] and [EEE]. This leaves us with the units [RR], [EEE], S, V, D to arrange, which is 5! ways.
The number of ways to arrange the Es together is \(\frac{6!}{2!} = 360\).
The probability is \(\frac{5!}{\frac{6!}{2!}} = \frac{120}{360} = \frac{1}{3}\).
