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Probability help

mello85

New member
Joined
Nov 17, 2009
Messages
3
Hello,

I've never been much of a math whiz so I figured I'd stop by here for some advice. I hope some of you can help, I would really appreciate it :)

Assume that I have the letters and numbers below separated into two piles, letters in one pile, numbers in the other. What are the odds of matching (for instance) H with 2, and D with 9? assuming I draw all of the items from both piles. What is the total number of possible combinations? 11^2? or 11^11?

A - 1
B - 2
C - 3
D - 4
E - 5
F - 6
G - 7
H - 8
I - 9
J - 10
K - 11

Thank you in advance! I look forward to hearing your replies. Please let me know if you need any more information, I've thought about it and I think that should be all the information you need :)
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, mello85!

The question is not stated clearly . . .


I have 11 letters in one pile and 11 numbers in another pile.
I draw an item from each pile, pairing the items.

(b) What is the total number of possible combinations? 11^2 or 11^11 ? . . . . neither

\(\displaystyle \begin{array}{cccccccccccc}\text{Arrange the letters in some order:} & A&B&C&D&E&F&G&H&I&J&K \\ \text{Below them, arrange the numbers:} &1&2&3&4&5&6&7&8&9&10&11 \end{array}\)

\(\displaystyle \text{We can see that there are: }\,11!\text{ ways to assign the numbers.}\)




(a) What is the probability of matching (for instance) H with 2, and D with 9 ?

Do you want exactly one match? . . . or exactly two matches? . . . or what?

Suppose we want the probability that H is paired with 2 and D is paired with 9.
. . There is 1 way that H and 2 are paired, and 1 way that D and 9 are paired,
. . and we don't care how the other 9 pairs are assigned: 9! ways.

\(\displaystyle \text{Hence: }\:p\bigg([H,2]\: \wedge \;[D,9]\bigg) \:=\:1\cdot1\cdot\frac{9!}{11!} \:=\:\frac{1}{110}\)

 

mello85

New member
Joined
Nov 17, 2009
Messages
3
Hello, mello85!

The question is not stated clearly . . .

I have 11 letters in one pile and 11 numbers in another pile.
I draw an item from each pile, pairing the items.

(b) What is the total number of possible combinations? 11^2 or 11^11 ? . . . . neither
Surely its more than 11 possible combinations. "A" can pair with "1-11", "B" can pair with "1-11", "C" can pair with "1-11" and so on, with just the first 3 letters there are already 33 possible combinations.

Do you want exactly one match? . . . or exactly two matches? . . . or what?
I'll try to put it in a different way, it's kind of difficult to explain. If I drew from both piles until there are none left, what are the chances that in a single drawing session I will pick H with 2 and D with 9. Think of it as the lotto, out of all the possibilities you have to match 2 sets to win in a single drawing. If I match H with 2, I assume my chances are pretty slim that I'll also match D with 9 in the same session.
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, mello85!

Surely its more than 11 possible combinations.

Who said the answer is "eleven"?
Did you read my explanation and answer?


Think of it as the lotto.
Out of all the possibilities, you have to match 2 sets to win in a single drawing.

\(\displaystyle \text{So we must match }H\text{ with }2\text{ and }D\text{ with }9\text{ in order to win the jackpot.}\)

\(\displaystyle \text{There are: }\:11!\text{ (eleven factorial) possible outcomes.}\)

\(\displaystyle \text{There is }1\text{ way to match }H\text{ with }2.\)
\(\displaystyle \text{There is }1\text{ way to match }D\text{ with }9.\)

\(\displaystyle \text{We don't care about the other 9 pairings.}\)
\(\displaystyle \text{There are: }\:9!\text{ possible pairings.}\)

\(\displaystyle \text{Hence, there are: }\:1\cdot1\cdot9!\text{ ways to get the two matches.}\)


\(\displaystyle \text{Therefore: }\;P(H\!=\!2 \:\wedge\: D\!=\!9) \;=\;\frac{9!}{11!} \;=\;\frac{1}{110}\)

 

mello85

New member
Joined
Nov 17, 2009
Messages
3
Excellent, thanks.

I wasn't sure what you were doing with the factorials there. I was getting some huge numbers, 362880/39916800 until I realized it was merely simplified.
I think I made things a bit more complicated than they really were. Thanks for breaking it down for me, I appreciate the help :)
 
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