Each permutation of the letters in THEORY is written on a slip of paper and tossed into a bowl. One slip is drawn at random, what is the probability that the letter T willappear in the last position?
Hello, Jigglypuff!
Yet another approach . . .
There are $\displaystyle 6! = 720$ permutations of the six letters.Each permutation of the letters in $\displaystyle THEORY$ is written on a slip of paper
and tossed into a bowl. One slip is drawn at random.
What is the probability that the letter $\displaystyle T$ will appear in the last position?
. . Hence, there are 720 slips of paper in the bowl.
If $\displaystyle T$ is last: .$\displaystyle \_\:\_\:\_\:\_\:\_\:T$
. . the other 5 letters can be arranged in $\displaystyle 5! = 120$ ways.
Hence, there are 120 slips with $\displaystyle T$ in the last position.
Therefore: .$\displaystyle P(\text{ T is last }) \;=\;\frac{120}{720} \;=\;\boxed{\frac{1}{6}}$