Math Help Forum

Math Help Forum Feed Site Feed

Go Back   Math Help Forum > Pre-University Math Help > Basic Statistics and Probability
Reply
 
Thread Tools Display Modes
  #1  
Old October 12th, 2009, 04:40 PM
Newbie
 
Join Date: Sep 2009
Location: Duluth, California
Posts: 14
Country:
Thanks: 3
Thanked 1 Time in 1 Post
Sarah- is on a distinguished road
Send a message via AIM to Sarah- Send a message via Skype™ to Sarah-
Post Help with high school algebra problem (some probability)

Keekerik is an imaginary land where the people have an interesting three-stage ritual for couples who want to get married. Wandalina and Gerik are in that situation so they go the the hope of Queen Katalana to perform this ritual. Permission for them to marry as soon as they wish depends on the outcome of the ritual.

The queen greets them and reaches into a box and pulls out six identical strings for the ritual. The queen hands the strings to Wandalina who holds them firmly in her fist. One end of each string is sticking out above and below her fists.

The queen steps to the side and Gerik is called forward. He ties two of the ends together above Wandalina's fist. Then he ties two other ends above her fist together. Finally, he ties the last two ends above her fist together. The six ends below Wandalina's fist are still hanging untied.

Now Queen katalana comes forward again. Although she was watching Gerik, she has no idea which string end below Wandalina's fist belongs to which end above. The queen does the final step and randomly picks two of the ends below and ties them together, then two more, and finally the last two. So Wandalina now has six strings in her first with three knots above and three knots below.

Whether Wandalina and Gerik will be allowed to marry right away depends on what happens when Wandalina opens her first. If the six strings form one large loop, then they will. Otherwise, they will be required to wait and reapat the ritual in six months.

Questions:

A. When Wandalina opens her first and looks at the strings, what combination of different size loops might there be?

B. What is the probability that the strings will form one big loop? In other words, what are the chances that Wandalina and Gerik will be able to marry right away?

C. What is the probability for each of the other possible combinations?

-----------------------------

I'm no math expert (that's why I'm asking you), but I guess this entire problem is mostly about probability. How should I go about answering these questions? I'm totally stuck.
Reply With Quote
Advertisement
 
  #2  
Old October 12th, 2009, 10:47 PM
Super Member

 
Join Date: May 2006
Location: Lexington, MA (USA)
Posts: 7,286
Thanks: 555
Thanked 4,649 Times in 3,712 Posts
Soroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond repute
Default

Hello, Sarah-!

Who assigned this problem?
The solution is far too long . . .


Quote:
Keekerik is an imaginary land where the people have an interesting three-stage ritual
for couples who want to get married. Wandalina and Gerik are in that situation
so they go the the hope of Queen Katalana to perform this ritual. Permission for them
to marry as soon as they wish depends on the outcome of the ritual.

The Queen reaches into a box and pulls out six identical strings for the ritual.
The Queen hands the strings to Wandalina who holds them firmly in her fist.
One end of each string is sticking out above and below her fists.

The Queen steps to the side and Gerik is called forward.
He ties two of the ends together above Wandalina's fist.
Then he ties two other ends above her fist together.
Finally, he ties the last two ends above her fist together.
The six ends below Wandalina's fist are still hanging untied.

Now Queen katalana comes forward again. Although she was watching Gerik,
she has no idea which string end below Wandalina's fist belongs to which end above.
The Queen does the final step and randomly picks two of the ends below
and ties them together, then two more, and finally the last two.
So Wandalina now has six strings in her first with three knots above and three knots below.

Whether Wandalina and Gerik will be allowed to marry right away depends on what happens
when Wandalina opens her first. If the six strings form one large loop, then they will.
Otherwise, they will be required to wait and repeat the ritual in six months.

Questions:

A. When Wandalina opens her first and looks at the strings,
what combination of different size loops might there be?
There are 225 possible outcomes, but only 4 configurations:
. . Three loops of two strings each.
. . Two loops, one with two strings, one with four strings.
. . Two loops, both with three strings.
. . One big loop, using all six strings.




Quote:
B. What is the probability that the strings will form one big loop?
Number the strings from 1 to 6.

Gerik can tie the upper ends in 15 different ways.

. . \begin{array}{c}(12, 34, 56) \\ (12,35,46) \\ (12,36,45) \end{array} \quad\begin{array}{c} (13,24,56) \\ (13,25,46) \\ (13,26,45) \end{array} . . \begin{array}{c}(14,23,56) \\ (14,25,36) \\ (14,26,35) \end{array}\quad \begin{array}{c}(15,23,46) \\ (15,24,36) \\ (15,26,34) \end{array} . . \begin{array}{c}(16,23,45) \\ (16,24,35) \\ (16,25,34) \end{array}

The Queen has the same 15 choices for tying the lower ends.

Hence, there are: .15 \times 15 \:=\:225 possible outcomes.



For any one of Gerik's choices, the Queen has 8 choices for forming one big loop.

For example, if Gerik chooses this combination (12,34,56)
. . the Queen can form one big loop with one of these eight choices:

. . . . \begin{array}{ccc}(13,25,46) &&(15,23,46) \\
(13,26,45) && (16,24,36) \\
(14,25,36) && (16,24,35) \\
(14,26,35) && (16,23,45) \end{array}

I believe the probability of one big loop is \frac{8}{15}



And that is all the work I'm willing to do for this problem . . .
.
Reply With Quote
The following users thank Soroban for this useful post:
Donate to MHF
  #3  
Old October 13th, 2009, 02:46 PM
Newbie
 
Join Date: Sep 2009
Location: Duluth, California
Posts: 14
Country:
Thanks: 3
Thanked 1 Time in 1 Post
Sarah- is on a distinguished road
Send a message via AIM to Sarah- Send a message via Skype™ to Sarah-
Default

Thanks for your suggestions! I'm still pretty lost in this one, but I'm trying to figure it out. If anyone else wants to chime in and give me assistance, it's greatly appreciated.
Reply With Quote
Reply

Tags
algebra, probability

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off
Forum Jump


All times are GMT -7. The time now is 11:45 PM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2009, Jelsoft Enterprises Ltd.
SEO by vBSEO 3.2.0 ©2008, Crawlability, Inc.
©2005 - 2009 Math Help Forum


Math Help Forum is a community of maths forums with an emphasis on maths help in all levels of mathematics.
Register to post your math questions or just hang out and try some of our math games or visit the arcade.