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Letters are chosen without replacement. I get it that if I was to choose, lets say the letter C, then my E(x)=(1/6)(1+2+3+4+5+6). Because I have an equal chance to choose the letters for C.

But when I have two O's in the word, why isn't E(x)=(1/6)(1+2+3+4)+(1/3)(1+2). The answer in the back of the book says 14/3. No matter how I format this E(x) I stated, I can't get it. I was hoping to just find the correct equation and be able to figure out why it works.

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  • $\begingroup$ Asks question at $T_0$, accepts answer at $T_0+90'$, asks for explanations at $T_0+120'$... $\endgroup$ – Did Mar 26 '14 at 5:30
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$P$(it takes 5 choices) = $P$(choosing a string of five letters containing two O's with an O at the end)

$$= \frac{\textrm{total number of such strings which are distinct}}{\textrm{total number of distinct strings of six letters}}$$

Total number of such strings which are distinct

= 4 (total number of ways of ordering the letters C,H,S,E)

= $4 \cdot 24 = 96, $

where the factor of $4$ corresponds to the the number of ways of positioning the first O among the first three letters chosen from C,H,S,E.

Total number of distinct strings of six letters = $\frac{6!}{2} = 360$.

So $P$(it takes five choices) = $\frac{96}{320} $.

Using the same method you can calculate the probabilites $P$(it takes $n$ choices) for $n \in \{2,3,4,5,6\}$.

You can then calculate the expectation using the formula:

$$E(X) = \sum_{n=2}^6 n P(\textrm{it takes $n$ choices}) \ . $$

General formula:

$P(X=n) = \frac{(n-2)! \cdot (6-n)! \cdot (n-1)}{320}$ for $n=2,3,4,5,6.$

The $(n-2)!$ corresponds to the non-O letters before the second O.

The $(6-n)!$ corresponds to the letters after the second O.

The $(n-1)$ corresponds to the placing of the first O among the letters preceding the second O.

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  • $\begingroup$ So, I tried reconstructing the summation equation, but I am not getting what I need. From what I gathered from your answer, P(n)=((x-1)(x-1)!)/360. So P(6)=5*5!/320 P(5) already done P(4) = 3*3!/320 and so on... $\endgroup$ – user1908772 Mar 26 '14 at 3:35
  • $\begingroup$ I have posted what I think is the general formula in the answer. $\endgroup$ – Frank Mar 26 '14 at 3:53
  • $\begingroup$ I have computed the sum using the app here: wolframalpha.com/widget/… $\endgroup$ – Frank Mar 26 '14 at 3:59

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