In an arrangement of 6 letters where each letter is selected at random from the alphabet (letters can be selected repeatedly).
What is the probability of getting the word 'ASSESS' if after selecting 6 letters you have the option to have the ordering of the selected letters to be randomly rearranged once?
Solution
To tackle this question my initial thought is to work out the probability of the following points (edited after incorporating Peter's response)
- Probability of picking correct letters with correct ordering
(1/26)x(1/26)x(1/26)x(1/26)x(1/26)x(1/26)
- Probability of picking the correct letters with incorrect ordering
Total outcomes = 26^6
Desired outcomes (including the outcome of correct ordering of letters) = (6!)/(4!x1!x1!) = 30
Desired outcomes probability (including the outcome of correct ordering of letters) = 30 / (26^6)
Desired outcomes probability minus probability of correct ordering of letters (point 1)
= Probability of picking the correct letters with incorrect ordering
= 29 / (26^6)
- Given point 2, the probability of rearranging the correct letters with incorrect order into correct ordering
Total permutations (discounting 'S' is repeated 4 times)
(6!/4!)
Therefore probability of the random rearrangement arranging the letters into correct ordering = 1/30
Final answer:
After working out all 3 points above, the final answer will be:
Probability of point 1 + (Probability of point 2 x Probability of point 3):
=[ 1/26^6 ] + [ (29/26^6) x (1/30) ]
= 59 / (26^6 x 30)
Best Answer
There are $$26^6$$ possibilities to draw $6$ letters. The number of ways to draw $4$ times $S$ , once $E$ and once $A$ is $$\frac{6!}{4!\cdot 1!\cdot 1!}=\frac{6!}{4!}=30$$
Since drawing the word exactly (including the order) does not count as a success, we have $29$ possibilities to be successful.
Hence, the probability of a success is $$\frac{29}{26^6}\approx 9.4\cdot 10^{-8}$$