#### dcowboys107

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- Thread starter dcowboys107
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Are you given an intended answer, if so what is it?

The reason I ask is that I find this a rather advanced question.

If one selects any four of the letters from MATHEMATICS there are 136 ways to do that. The is the coefficient of x<SUP>4</SUP> in the expansion of \(\displaystyle \left( {1 + x + x^2 } \right)^3 \left( {1 + x} \right)^5\).

Only one of those is <M,A,T,H>. Answer 1/136.

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Thanks, I'll relay the answer back to you during class tomorrow. This is from Algebra II.

shuffle them and give out 4 of them: were the 4 cards numbered 1,2,3 and 4 (any order)?

So the probability of the 1st card being in the range is 7/11.

I get 3/125 as probability.

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The answers are all wrong, I checked with my teacher. She still won't tell me the answer.

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There are eight distinct letters is the word.

Randomly select four letters from the list above.

Here is a list of what can happen.

All four letters are distinct: This can happen in \(\displaystyle \L

\left( \begin{array}{c}

8 \\

4 \\

\end{array} \right) = 70\) ways.

One of the letters can appear twice, such as <MAAT> or <EHTT>.

This can happen in \(\displaystyle \L

\left( \begin{array}{c}

3 \\

1 \\

\end{array} \right)\left( \begin{array}{c}

7 \\

2 \\

\end{array} \right) = 63\) ways.

We can chose the repeated letter in 3 ways and the other two from 7.

Two letters can appear twice: <MMAA>, <MMTT>, or <AATT>.

AS you can see there are 3 ways this can happen.

The sum is 136. See my edited post above (the real mathematics is done there.)

This means that there are a total of 136 4-multisets such that as <MATT> and <MATH> is only one of those.

Thus the answer is 1/136 regardless.

The above is the answer to this question.

If the question read: What is the probability that if you choose 4 letters at random from the word MATHEMATICS one at a time

The answer to that question is \(\displaystyle \L

\frac{2}{{11}} \cdot \frac{2}{{10}} \cdot \frac{2}{9} \cdot \frac{1}{8}\)

pka said:"Thus the answer is 1/136 regardless."

Disagree; I make it in vicinity of 1/41.

I ran simulations and kept getting ~24,400 per million; 99.9% sure it's accurate.

However, I'm not sure how to prove that mathematically...but I'll find a way!

"If the question read: What is the probability that if you choose 4 letters at random from the word MATHEMATICS one at a timethey will spell MATH in the order they are drawn?

The answer to that question is \(\displaystyle \L

\frac{2}{{11}} \cdot \frac{2}{{10}} \cdot \frac{2}{9} \cdot \frac{1}{8}\)

AGREE!

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\(\displaystyle \L

\begin{array}{l}

\frac{2}{{11}} \times \frac{2}{{10}} \times \frac{2}{9} \times \frac{1}{8} \times (4!) = .0242424 \\

\frac{1}{{41}} = .0243 \\

\end{array}\)

My interest is in counting theory and not so much probability.

I know that there are 136 different samples: multi-sets.

It occurs to me that they may have a different probability of occurring.

5/7 3/6 1/5 = 1/14 : "3" pulled last

5/7 4/6 2/5 = 4/21 : "3" pulled first

5/7 3/6 2/5 = 1/7 : "3" pulled second

Each of these 3 possible outcomes has its own probability of occuring;

that's what I'm not sure how to calculate; making 'em x,y,z, then:

P = (1/14)x + (4/21)y + (1/7)z

Hmmm...well :?:

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The probability of ‘123’ in that order is \(\displaystyle \LDenis said:Using a simpler example: 1,1,2,2,3,4,5 ; we want 1,2,3 in any order.

\frac{2}{7} \times \frac{2}{6} \times \frac{1}{5}\) .

But the string can appear in (3!)

So the probability of 1,2,3 in any order is \(\displaystyle \L

\left( 3! \right)\left( {\frac{2}{7} \times \frac{1}{6} \times \frac{1}{5}} \right)\)

The probability of ‘225’ in that order is \(\displaystyle \L

\frac{2}{7} \times \frac{1}{6} \times \frac{1}{5}\) .

The string can appear in 3 ways.

So the probability of 2,2,5 in any order is \(\displaystyle \L

\left( 3 \right)\left( {\frac{2}{7} \times \frac{1}{6} \times \frac{1}{5}} \right)\)