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Suppose that the word "mississippi" is written on a piece of paper. If we cut this piece of paper into 11 smaller pieces of paper, each containing exactly one letter, and then place these letters in a bag, what is the probability that if we select four numbers at random with no replacement, the first letter will be "m", the second letter "i", the third letter "s", and the fourth letter is "s" (so that we form the word "miss")?

At this point, I know that computing ${}_{11}P_4$ will give us the total number of ways we can form $4$-letter words. Would I have to divide ${}_{11}P_4$ by $4!\cdot 4! \cdot 2!$ to get the total number of ways to uniquely spell out "miss". Then, I could find the probability by dividing $1$ by the total number of ways found?

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At each stage calculate the probability of picking the required letter and multiply. The M has probability $\frac1{11}$ of being picked, the I $\frac4{10}$, the first S $\frac49$ and the second S $\frac38$. The final answer is $\frac1{165}$.

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  • $\begingroup$ @K.M $\frac{1}{165}$ is the correct answer, not $\frac{1}{990}$ $\endgroup$ – user71207 Jan 14 at 6:55
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One way to do this is to find the number of ways to rearrange the letters such that the first four letters are the same (i.e. the number of ways to rearrange the letters of "issippi"), and divide this by the number of ways to rearrange the letters of "mississippi." This can be calculated using multinomial coefficents as follows, where the top number is the number of letters in the word, and the bottom numbers are the counts of each unique letter.

$$\frac{ \begin{pmatrix} 7\\ 2,2,3 \end{pmatrix} }{ \begin{pmatrix} 11\\ 1,2,4,4 \end{pmatrix}} = \frac{210}{34650} = \frac{1}{165} $$

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The TOTAL number of ways you could select 4 letters out of 11 would be 11P4.

So, the sample space (denominator) of the probability would be 11P4.

Sample space = $7920$

When we choose miss out of mississippi we could choose from one m and two sss and 4 is.

So there would be 4P2 = $12$ ways where the ss part of miss could be chosen. And there are 4 is to choose from. And only one m to choose from.

So the total number of ways miss ALONE can be selected = $1*4*12 = 48$ ways

So, therefore, the probability to spell out miss would be $\frac{8}{7920}=\frac{1}{165}$

Comment if my answer doesn't satisfy you.

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  • $\begingroup$ Why are there two ways for the ss to be together? $\endgroup$ – K.M Jan 14 at 6:51
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    $\begingroup$ @K.M there are actually ${4\choose2} = 6$ ways for the SS to be together $\endgroup$ – user71207 Jan 14 at 6:57
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    $\begingroup$ Sorry it should actually be $4P2 = 12$. Then this gives $1\cdot12\cdot4 = 48$. Hence the probability is $\frac{48}{7921} = \frac{1}{165}$. $\endgroup$ – user71207 Jan 14 at 7:06
  • $\begingroup$ yes @user71207 i'll edit it $\endgroup$ – HyperFluxx Jan 15 at 7:22

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