$Combinatorical Drawing$

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$Combinatorical Drawing$

If there is a pool of 2 Ms, 2 As, 2 Ts, 2Hs, and 1 X, Y, Z, how would you find the probability that if you drew 7 letter cards that you could spell Math from what you drew?

Combinatorics Probability

Matchmaticians Alumnus

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To spell Math, we first have to pick one letter M, then one letter A, then T and finally H.

The probability that the first pick is M is 2/11, since there are two letter M's available and 11 letters in total.

Once we have picked M, the probability of picking an A is 2/10, since there are two letter A's and the pool of letters now only has 10 letters, since we have already picked one M out.

Once we have picked M,A, the probability of picking a T is 2/9, following the same logic.

Finallky, after picking M,A,T, the probability of picking H is 2/8.

The final answer would be: (2/11)*(2/10)*(2/9)*(2/8)

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Edit:

If ordering doesn't matter, but only the possibility of spelling MATH out of the picked letters, we could solve it like this:

From the two M's we need one. That gives us a factor of 2C1.
From the two A's we need one. That gives us a factor of 2C1.
From the two T's we need one. That gives us a factor of 2C1.
From the two H's we need one. That gives us a factor of 2C1.
From the remaining 7 letters, we need to pick other 3 to complete a draw of 7. That gives us a factor of 7C3.

And the total possible ways of choosing 7 letters out of 11 is 11C7

The final answer is:

(2C1)*(2C1)*(2C1)*(2C1)*(7C3)/(11C7)

Mathe

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Matchmaticians Alumnus

0

You have given the probability of drawing a M, followed by an A, followed by a T, followed by an H. I'm looking for a technique that likely uses combinations and inclusion/exclusion to compute the probability. Something like: total of 7 letter combinations with at least 1 M, A, T, and H over 11 choose 7.
- Mathe
  
  0
  
  I'm afraid I don't understand your question, but I think it's different from what is stated in the problem.
Matchmaticians Alumnus

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So there are 11 letters and you draws 7 of them. What is the probability that you can spell Math from what you drew?
- Mathe
  
  0
  
  I added a new response in the case that the order of the picks is not relevant.
Matchmaticians Alumnus

0

On second thought, I think you are overcounting. If you drew a D1 in the 2C1 you could draw a D2 in the 7C3, which is fine, but you could also draw a D2 in the 2C1 and a D1 in the 7C3, which is just a reordering of the prior event.
- Mathe
  
  0
  
  What do you mean by a D1 and a D2? I think you mean an 'M', for example. I must draw an 'M' at some point. The first M I draw is counted in the 2C1 factor. A second M would be counted in the 7C3.
Matchmaticians Alumnus

0

Yes, meant M. I contend that the numerator, which represents the number of ways to have M A T H and 3 other letters is too large. The reason for this is, because, say you let M1 a 2C1, and M2 be in the 7C3. This is a valid draw. However, if in the future, if you have an M2 in the 2C1, and a M1 in the 7C3, with the other 5 characters the same, this event would be counted again with the method you presented.
- Mathe
  
  0
  
  It would also be counted twice in the denominator.
Matchmaticians Alumnus

0

It wouldn't be counted twice in the denominator because the denominator is counting unordered events. If you did a permutation, then sure, but the numerator would need to be adjusted.

$Combinatorical Drawing$

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