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?

Printable View

- May 18th 2008, 05:15 PMJigglypuffPermutation of theory
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?

- May 18th 2008, 05:31 PMUnenlightened
THEORY... Six letters.

Only one can come last at any given time.

There's a one in six chance that this will be 'T'.

So: 1/6.

That's probably the simplest way of looking at it...

You could also say that you had:

5/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1/1 = 1/6.

I think... - May 18th 2008, 07:59 PMSoroban
Hello, Jigglypuff!

Yet another approach . . .

Quote:

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: .$\displaystyle P(\text{ T is last }) \;=\;\frac{120}{720} \;=\;\boxed{\frac{1}{6}}$