Probability question about letters Let's say we had the word STATISTICS, and it was hanging on a wall, then two letters in the word blew off, and somebody found them. What are the chances they put them in the correct order?
here is what I have. 
A = letters are put in order
I = letters are identical
P(A given I) = 1
P(A given I compliment) = 1/2
I want to use the total probability theorem, but I have a hard time grasping what P(I) could be, I'm sure it would be 1 / 10 NCR something, but I'm not sure what to put. 
Any ideas?
Here is a solved version of this problem.
Word = CANAL
A = words are in the right order
I = words are identical
P(A given I ) = 1
P(A given I^c) = 1/2
P(I) = 1/5 choose 2
P(A) = 1*1/10 + 1/2*9/10
by the total probability theorem.
 A: This was my first (incorrect) interpretation of the OP's question (I've added another answer which I believe has the correct interpretation).
Two letters fall off. Assuming the person replacing them knows how to spell, if different letters fall they will certainly be replaced correctly. If the same letters fall off, they must be replaced in the same order in which they were originally (correct spelling AND correct original position). Under these assumptions, the answer is:
$P(A) = P(different\,letters\,fell\,off) + P(same\,letters\,fell\,and\,replaced\,correctly)$
We have 4 bins (S, T, A, I, C) with (3, 3, 1, 2, 1) letters each. 
$P(same\,letters\,fell\,off) = P(2\,S\,fell) + P(2\,T\,fell) + P(2\,A\,fell) + P(2\,I\,fell) + P(2\,C\,fell)$
$= 3 / 10 * 2 / 9 + 3 / 10 * 2 / 9 + 0 + 2 / 10 * 1 / 9 + 0 = 14 / 90$
$P(different\,letters\,fell\,off) = 1 - P(same\,letters\,fell\,off) = 76 / 90$
$P(same\,letters\,fell\,and\,replaced\,correctly) = P(same\,letters\,fell\,off) * P(replaced\,correctly) = 14 / 90 * 1 / 2 = 14 / 180$
$P(A) = 76 / 90 + 14 / 180 = 166 / 180$
A: So here goes another answer using a different interpretation of the question: two letters fall off. They will be replaced randomly. A correct replacement is such that the spelling of the word is unchanged.
$P(A) = P(same\,letters\,fell\,off) + P(different\,letters\,fell\,and\,replaced\,correctly)$
We have 4 bins (S, T, A, I, C) with (3, 3, 1, 2, 1) letters each. 
$P(same\,letters\,fell\,off) = P(2\,S\,fell) + P(2\,T\,fell) + P(2\,A\,fell) + P(2\,I\,fell) + P(2\,C\,fell)$
$= 3 / 10 * 2 / 9 + 3 / 10 * 2 / 9 + 0 + 2 / 10 * 1 / 9 + 0 = 14 / 90$
$P(different\,letters\,fell\,off) = 1 - P(same\,letters\,fell\,off) = 76 / 90$
$P(different\,letters\,fell\,and\,replaced\,correctly) = P(different\,letters\,fell\,off) * P(replaced\,correctly) = 76 / 90 * 1 / 2 = 76 / 180$
$P(A) = 14 / 90 + 76 / 180 = 104 / 180$
Using the word CANAL, buckets are (C, A, N, L) with sizes (1, 2, 1, 1).
$P(same\,letters\,fell\,off) = P(2\,C\,fell) + P(2\,A\,fell) + P(2\,N\,fell) + P(2\,L\,fell)$
$= 0 + 2 / 5 * 1 / 4 + 0 + 0 = 2 / 20$
$P(different\,letters\,fell\,off) = 1 - P(same\,letters\,fell\,off) = 18 / 20$
$P(different\,letters\,fell\,and\,replaced\,correctly) = P(different\,letters\,fell\,off) * P(replaced\,correctly) = 18 / 20 * 1 / 2 = 18 / 40$
$P(A) = 2 / 20 + 18 / 40 = 22 / 40 = .55$
which agrees with your book answer.
