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The original problem is:

"You choose a letter at random from the word Mississippi eleven times without replacement. What is the probability that you can form the word Mississippi with the eleven chosen letters? Hint: it may be helpful to number the eleven letters as 1,2,...,11."

I've looked through other pages which tell me the answer should be: 2.88*10^-5

Yet my book tells me it's ${11 \choose 4}*{7 \choose 4}*{3 \choose 2}*4^4*4^4*2^2*(1/(11^{11}))$ = .0318...

Is my book wrong? How would you approach this question?

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    $\begingroup$ If you choose 11 letters from "Mississippi" without replacement, then you are guaranteed to have chosen all 11 letters of the word and the probability is 1 that you can "form the word" again with those letters.. Did you mean "with replacement"? Or did you mean, what is the probability that the sequence of choices without replacement will spell "Mississippi"? Please clarify. $\endgroup$ – Rob Arthan Aug 7 '18 at 1:14
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    $\begingroup$ By way of additional background, I found this exercise in "Understanding Probability, Third Edition" by Henk Tijms, Problem 7.9 on page 236, and yes, the problem statement does say "without replacement", but that is clearly a mistake in the book, because the problem statement does not agree with the solution given in the back of the book. $\endgroup$ – awkward Aug 8 '18 at 12:40
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Your book is essentially asking what is the probability that picking $11$ letters from Mississippi WITH replacement gives you $11$ letters that can be rearranged to form Mississippi.

I agree with your interpretation, and the way you did it is absolutely consistent, but that's how the book (badly) interprets it.


One way is to simply this is first find the probability that you can get SSSSIIIIPPM which is $\left(\frac{4}{11}\right)^4\left(\frac{4}{11}\right)^4\left(\frac{2}{11}\right)^2\left(\frac{1}{11}\right)^1$. That can be rearranged into MISSISSIPPI. There are $\frac{11!}{4!4!2!1!}$ different permutations of MISSISSIPPI. So we have to find the probability of getting any one of those permutations, which is a $\frac{11!}{4!4!2!1!}\left(\frac{4}{11}\right)^4\left(\frac{4}{11}\right)^4\left(\frac{2}{11}\right)^2\left(\frac{1}{11}\right)^1 = \boxed{0.031837}$ chance.

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    $\begingroup$ OMG! Thank you! I've been trying to understand this for so long now! I would upvote your answer, it's just that I'm new and the site won't record my upvotes until I have "15 reputation" $\endgroup$ – BigBear Aug 7 '18 at 2:21
  • $\begingroup$ But the quotation says "WITHOUT replacement"? I don't see how we can answer the question as it stands. $\endgroup$ – Rob Arthan Aug 7 '18 at 22:46
  • $\begingroup$ I looked it up. The author meant "with replacement" kinda annoying but I supposed it's a simple human error $\endgroup$ – BigBear Aug 8 '18 at 0:48
  • $\begingroup$ @RobArthan it was an error $\endgroup$ – Saketh Malyala Aug 8 '18 at 4:03
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In the word Mississippi there are $4$ s, $4$ i, $2$ p.

So, the number of permutations $=\dfrac{11!}{2!\times4!\times4! }=32650$

The probability that you get the word Mississippi with eleven chosen letters is $$\dfrac{1}{\dfrac{11!}{2!\times4!\times4!}}=\dfrac{2!\times4!\times4!}{11!}=\dfrac{1}{32650}=2.88\times10^{-5}$$

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I am assuming the problem is about letters taken from Mississippi with replacement, not without replacement as stated in the OP. The following solution is similar to one previously posted, but with additional justification as to why the book solution is the correct choice.

In problems like this it is important to decide on the probability space, which often amounts to finding a collection of equally likely outcomes, then counting those that are favorable in one sense or another. Here, we want to view Mississippi as if all the letters were distinct, perhaps by applying subscripts to the letters, so the starting word is $$M_1 I_1 S_1 S_2 I_2 S_3 S_4 I_3 P_1 P_2 I_4$$ Then 11 letters taken from that word with replacement can be arranged in $11^{11}$ distinct ways, all of which we assume are equally likely.

The favorable outcomes we want to count are those sequences which contain 1 M, 4 Is, 4 Ss, and 2 Ps. The number of sequences of 1 M, 4 Is, 4 Ss, and 2 Ps without the subscripts is $$\frac{11!}{1! \;4! \;4! \;2!}$$ but we need to count the number of sequences with subscripts in order to be consistent with our definition of equally likely possibilities. Each $M$ can be subscripted in only $1$ way, each $I$ can be subscripted in $4$ ways, etc., so the number of favorable outcomes is $$\frac{11!}{1! \;4! \;4! \;2!}\cdot 1 \cdot 4^4 \cdot 4^4 \cdot 2^2$$ Divide the number of favorable outcomes by $11^{11}$ to find the probability of a favorable outcome. $$\frac{11!}{1! \;4! \;4! \;2!}\cdot \frac{1 \cdot 4^4 \cdot 4^4 \cdot 2^2}{11^{11}} = 0.0318364$$

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n=11 objects

We need to find num of linear arrangements

Types: M:1 I:4 S:4 P:2

11!/(1!4!4!2!)=(11.10.9.9.8.7.6.5.4.3.2.1)/(1)(4.3.2.1)(4.3.2.1)(2.1)

easy to simplify the fraction after expanding the factorials.

(69300)5=34,650 ways of writing MISSISSIPPI

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