probability

dcowboys107

New member
What is the probability that if you choose 4 letters at random from the word MATHEMATICS that you will select 4 letters that spell the word MATH?

galactus

Super Moderator
Staff member
How many letters does MATHEMATICS have?.
How many letters are you choosing?.
How many times does the M appear?. The A?. The T?. The H?.

pka

Elite Member
What course is this question from?
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.

dcowboys107

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

Denis

Senior Member
I see that the same as having 11 cards with numbers 1,1,2,2,3,3,4,5,6,7,8;
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.

dcowboys107

New member
the anwers is wrong

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

pka

Elite Member
M A T H E M A T I C S
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.
What is the probability that if you choose 4 letters at random from the word MATHEMATICS that you will select 4 letters that spell the word MATH?

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

The answer to that question is $$\displaystyle \L \frac{2}{{11}} \cdot \frac{2}{{10}} \cdot \frac{2}{9} \cdot \frac{1}{8}$$

However, that is not the question that was asked!

Denis

Senior Member
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 time they 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!

pka

Elite Member
Dennis, you may be correct. Note
$$\displaystyle \L \begin{array}{l} \frac{2}{{11}} \times \frac{2}{{10}} \times \frac{2}{9} \times \frac{1}{8} \times (4!) = .0242424 \\ \frac{1}{{41}} = .0243 \\ \end{array}$$

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.

Denis

Senior Member
Using a simpler example: 1,1,2,2,3,4,5 ; we want 1,2,3 in any order.

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 :?:

pka

Elite Member
Denis said:
Using a simpler example: 1,1,2,2,3,4,5 ; we want 1,2,3 in any order.
The probability of ‘123’ in that order is $$\displaystyle \L \frac{2}{7} \times \frac{2}{6} \times \frac{1}{5}$$ .

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

Denis

Senior Member
yes, yes, yes...of course! Thanks, pka.

"Edit" your "1/6"'s to 2/6 : typoes, right?

So the MATH one is 4! (2/11) (2/10) (2/9) (1/8) : deceptively easy :shock: