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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)

  1. 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)

  1. 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)

  1. 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)

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    $\begingroup$ Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? $\endgroup$ – 5xum Oct 2 '17 at 12:55
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    $\begingroup$ Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. $\endgroup$ – 5xum Oct 2 '17 at 12:55
  • $\begingroup$ As a way to get started: try a smaller collection. Try, say, $BOB$ with the same sort of rules. $\endgroup$ – lulu Oct 2 '17 at 12:57
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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 $$\frac{29}{26^6}\approx 9.4\cdot 10^{-8}$$

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  • $\begingroup$ Thanks Peter, I am not sure if I have over complicated the question but it seems this answer only address the probability of achieving ASSESS by selecting the right characters and the right order of characters only? $\endgroup$ – jacdam333 Oct 2 '17 at 13:16
  • $\begingroup$ @jacdam333 I considered that the order doesn't matter. The $30$ is the number of draws leading to the combination $SSSSAE$ without considering the order. We are successful, if we draw one of the $30$ draws. Successfull draws are, for example $ASSESS$ or $ESASSS$ $\endgroup$ – Peter Oct 2 '17 at 13:21
  • $\begingroup$ If we were not allowed to rearrange the letters, the probability of a success would be $$\frac{1}{26^6}$$ because then only one draw would be successful. $\endgroup$ – Peter Oct 2 '17 at 13:24
  • $\begingroup$ Thanks Peter, I have incorporated your working into my answer above. Please feel free to comment if I have made any mistakes. $\endgroup$ – jacdam333 Oct 2 '17 at 13:46
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    $\begingroup$ Drawing ASSESS including order do count as success. That is why I have added the probability of achieving point 1 into the final probability. Whereas based on the question, drawing the correct characters but incorrect order, you will have an opportunity to randomly rearrange these letters once. Therefore probability of point 2 and 3 is required $\endgroup$ – jacdam333 Oct 2 '17 at 13:53

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