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► » Romeo and Juliet starring YOU in happy-ending edition with your picture on cover! Wedding Anniversary Gifts and Unique Romantic Present Ideas - a customized romance novel! HOME : BOOKS : -- CORPORATE GIFTS : Kids : GIFT BASKETS : TESTIMONIALS : FAQ : CONTACT -- -- Call us toll free 1-866-660-1109 -- Romeo and Juliet - the happy ending edition! Starring YOU and a special someone as Romeo and Juliet! -- The ultimate romantic, wedding or anniversary gift — now available in a personalized "happy ending" edition, with optionally your pictures on the cover! It's the way Romeo and Juliet should have been - true love with a personal twist! Standard cover Photo cover Price: $29.95 “There could not have been a more perfect 20th wedding anniversary present. Thanks so much.” A.H., Studio City, CA Click here to read what our customers have been saying... About Customizations tell a friend Photos Gift Ideas Romeo and Juliet, Shakespeare's classic drama of passionate love—now releasedas a unique personalized romance with YOU playing the part of the famous lovers. A novel wedding,romantic or anniversary gift . Picture the thrill of classic lines personalized with you in them: "Oh Romeo, Romeo Brad, Brad. Wherefore art thou Romeo Brad?" "But, soft! What light through yonder window breaks? It is the east, and Juliet Helenis the sun." What's more, if you choose the Happy Ending Version a new scene is added withan unexpected plot twist — the lovers live happily ever after! A short scene is added after ActV Scene III. It turns out the apothecary's poison didn't work and Romeo survives, and Juliet's stabbingof herself merely made her pass out. (With sincere apologies to William Shakespeare, Mercutio and Tybalt!) Romeo : Juliet's true love Montague : Romeo's family, involved in a bitter feud with the Capulet clan (Juliet'sfamily) Juliet : Romeo's true love Capulet : Juliet's family, involved in a bitter feud with the Montague clan (Romeo'sfamily) Verona : The city where most of the story takes place Happy Ending Version The new scene is available in the following versions, whereRomeo and Juliet survive: Classic version ACT V, SCENE IV. IN THE SEPULCHRE. [Romeo and Juliet awaken, rubbing their eyes] Romeo What uncommon commotion stirs these folk? Ah, blessed apothecary, whose potion miss'dits mark! Juliet And perhaps 'twas the keenness of mine love that hath dulled the dagger's blade. Romeo What sayest thou we hasten to Verona? Juliet Come, prince, love, husband, shining angel! Let's leave this cold sepulchre for Verona'swarm embrace. [Exeunt Romeo and Juliet hand in hand] Irreverent version ACT V, SCENE IV. IN THE SEPULCHRE. [Romeo and Juliet awaken, rubbing their eyes] Romeo What the heck was that big scene all about? Juliet Who knows? I just passed out for a second and everybody's losing it. Luckily thedagger wasn't sharp. Romeo And the apothecary screwed up big-time! What do you say we head home? Juliet Sounds like a plan, my medieval man! [Exeunt Romeo and Juliet hand in hand] Your privacy is assured: once the email is sent we do not retain any names or email addresses Email To Subject Message text Your name Your email Additional photos coming soon! Anniversary Gifts Wedding Gifts Birthday Presents Romantic Gifts Christmas Gifts Graduation Gifts Valentine Gifts Listed below are both the modern and traditional anniversary gift names Anniversary number Traditional Gift name Modern Gift name First Paper Clocks Second Cotton China Third Leather Crystal & Glass Fourth Fruit & Flowers Appliances Fifth Wood Silverware Sixth Candy & Iron Wood Seventh Wool & Copper Desk Sets Eighth Bronze & Pottery Linens & Lace Ninth Pottery / Willow Leather Tenth Tin & Aluminum Diamond Jewelry Eleventh Steel Fashion Jewelry Twelfth Silk & Linen Pearls Thirteenth Lace Textiles / Furs Fourteenth Ivory Gold Jewelry Fifteenth Crystal Watches Twentieth China Platinum Twenty-Fifth Silver Silver Thirtieth Pearl Diamond Thirty-Fifth Coral Jade Fortieth Ruby Ruby Forty-Fifth Sapphire Sapphire Fiftieth Gold Gold Fifty-Fifth Emerald Emerald Sixtieth Diamond Diamond -- Price: $29.95 Did you know? Select the custom photo option for the ultimate in personalization- a picture of YOU on the cover! (Details...) Every book comes with a customized cover with your name on it A dedication message (upto a page) is included at no extra cost -- Shipping Basic shipping to USA/Canada: $4 Express: $9 (More details...) -- -- Titles Romance - Romeo and Juliet Adventure - Zorro -- - Robin Hood - NEW! - The JungleBook - Moby Dick Fantasy - The Wizard of Oz - Alice in Wonderland Mystery - The Adventuresof Sherlock Holmes - The Memoirs of Sherlock Holmes - The Hound of the Baskervilles - Early Sherlock Holmes Drama - A Christmas Carol - Anne ofGreen Gables - Anne of Avonlea - Anne of the Island Horror - Dracula - Dr. Jekyll & Mr. Hyde -- HOME : POLICIES : DEALERS & AFFILIATES : JOBS : CONTACT : RESOURCES Click to verify Shop with confidence! Customized Classics is a certified InternetSecure Merchant ---



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Birthday Present

Math Forum: Ask Dr. Math FAQ: The Birthday Problem -- Ask Dr. Math: FAQ The Birthday Problem Dr. Math FAQ || Classic Problems || Formulas || Search Dr. Math || Dr. Math Home Suppose you flip a coin and bet that it will come up tails. Since you are equally likely to get heads or tails, the probability of tails is 50%. This means that if you try this bet often, you should win about half the time. What if somebody offered to bet that at least two people in your math class had the same birthday? Would you take the bet? This question is more complicated than flipping a coin, because the chance of finding two people with the same birthday depends on the number of people you ask. If there were only one other person in your math class, you might be surprised to find out that she had the same birthday as you. If there were a pair of people with the same birthday in a class of 366 people, would you still be surprised? How large must a class be to make the probability of finding two people with the same birthday at least 50%? Let's forget about leap year when we solve this problem (no February 29 birthdays!) This way, we can assume that a year is always 365 days long. We'll start by figuring out the probability that two people have the same birthday. The first person can have any birthday. That gives him 365 possible birthdays out of 365 days, so the probability of the first person having the "right" birthday is 365/365, or 100%. The chance that the second person has the same birthday is 1/365. To find the probability that both people have this birthday, we have to multiply their separate probabilities. (365/365) * (1/365) = 1/365, or about 0.27%. Now, what about three people ? The chance of the first and second person sharing a birthday is still 1/365. The first and third person might share a birthday instead. The probability of that is 1/365 as well. But what if the second and third person shared a birthday? And what if all three of them had the same birthday? Things are getting complicated fast. Four or five people would be even messier. Is there a simpler way? To solve the birthday problem, we need to use one of the basic rules of probability: the sum of the probability that an event will happen and the probability that the event won't happen is always 1. (In other words, the chance that anything might or might nothappen is always 100%.) If we can work out the probability that no two people will have the same birthday, we can use this rule to find the probability that two people will share a birthday: P(event happens) + P(event doesn't happen) = 1 P(two people share birthday) + P(no two people share birthday) = 1 P(two people share birthday) = 1 - P(no two people share birthday). So, what is the probability that no two people will share a birthday? Again, the first person can have any birthday. The second person's birthday has to be different. There are 364 different days to choose from, so the chance that two people have different birthdays is 364/365. That leaves 363 birthdays out of 365 open for the third person. To find the probability that both the second person and the third person will have different birthdays, we have to multiply: (365/365) * (364/365) * (363/365) = 132 132/133 225, which is about 99.18%. If we want to know the probability that four people will all have different birthdays, we multiply again: (364/365) * (363/365) * (362/365) = 47 831 784/ 48 627 125, or about 98.36%. We can keep on going the same way as long as we want. A formula for the probability that n people have different birthdays is ((365-1)/365) * ((365-2)/365) * ((365-3)/365) * . . . * ((365-n+1)/365). If you know permutation notation, you can write this formula as (365_P_n)/(365^n). That's the same as 365! / ((365-n)! * 365^n). We've made some progress, but we still haven't answered the original question: how large must a class be to make the probability of finding two people with the same birthday at least 50%? We know that the probability of finding at least two people with the same birthday is 1 minus the probability that everybody has a different birthday, and we know how to find the probability that everybody has a different birthday for any number of people. The easiest way to find the right class size is to use a calculator to try different numbers in the formula. It turns out that the smallest class where the chance of finding two people with the same birthday is more than 50% is... a class of 23 people . (The probability is about 50.73%.) From the Dr. Math archives: Probability Theory: Coincidental Birthday Probability of the Same Birthday within a Group Birthday Probabilities Three Share a Birthday The Birthday Problem; Queuing at a Bank Birthday Probability, Class of 25 One Person of Seven Born on Monday Odds of Left-Handedness in a Group From the Web: The Birthday Problem: A short lesson in probability , George Reese A Java applet that you can use to test different class sizes (it works better with small classes) and graphs of the probability for different numbers of people. The Law of Small Errors , Keith Devlin The birthday problem, and related questions - what's the probability that someone will have your birthday? Birthday Surprises, Ivars Peterson Birthday Problem, Eric Weisstein's World of Mathematics Coincidence, Alexander Bogomolny How to Read Mathematics, Shai Simonson and Fernando Gouveau This article uses an explanation of the birthday problem as an example. An Introduction to Mathematica and the "Birthday Problem," Louie Beuschlein For a general review of probability: Probability, Dr. Math FAQ Probability in the Real World, Dr. Math FAQ - Ursula Whitcher, for the Math Forum Submit your ownquestion to Dr. Math [ Privacy Policy ] [ Terms of Use ] Math Forum Home || Math Library || Quick Reference || Math Forum Search Ask Dr. Math ® © 1994-2005 The Math Forum http://mathforum.org/dr.math/



Teacher Gift Pack Mr.

Congratulations to the winners of the Monthly Giveaway Congratulations to past Giveaway Winners: August 2005: Ms. Margaret E. Blood, Bridgewater, NJ $100 Target Gift Card Ms. Luanna M. Dorsett, Mount Gilead, NC $100 Staples Gift Card Ms. Leslie L. Gates, Waynesboro, PA $100 Staples Gift Card Ms. Mae G. Martin, Gainesville, GA $100 Target Gift Card Ms. Dianna K. Wright, Terre Haute, IN $100 Staples Gift Card July 2005: Ms. Gail L. Lyons, Nutley, NJ Bose Wave Radio June 2005: Ms. Gina Byrd, Knoxville, TN Weber Charcoal Grill with accessories Ms. Dawn M. Dougherty, Barnegat, NJ Weber Charcoal Grill with accessories Ms. Maureen Tarantino, Fair Lawn, NJ Weber Charcoal Grill with accessories May 2005: Ms. Chaerea Denning, Nashville, TN Alamo 7-Day Car Rental Certificate Ms. Nancy Andorko, Delran, NJ Alamo 7-Day Car Rental Certificate Ms. Deborah Kropf, Kendall Park, NJ Alamo 4-Day Car Rental Certificate Ms. Mary Cash, Cushing, OK Hertz $100 Car Rental Certificate Ms. Dawn Carpenter, Spokane, WA Hertz $100 Car Rental Certificate April 2005: Mr. Glynne Jones, Katy, TX “NEA Getaway” Vacation Package, 4-Day Alamo Car Rental, and Luggage March 2005: Ms. Jane Barrett, Champaign, IL Classic Gift Basket Ms. Christina Elliott, High Ridge, MO Classic Gift Basket Mr. Anthony McIver, Mc Rae, GA Classic Gift Basket Ms. Stephanie Yusman, Ballwin, MO Classic Gift Basket February 2005: Mrs. Kay Roberts, Red Springs, NC 4-Day Trip to Washington, DC January 2005: Ms. Kathleen Chirpich, Overland Park, KS $200 U.S. Savings Bonds Ms. Sandra Krupa, Bentleyville, PA $200 U.S. Savings Bonds Ms. Mary Stone, Lake St. Louis, MO $200 U.S. Savings Bonds December 2004: Ms. Maria Ammerata, Sayreville, NJ Deluxe Reference Set Ms. Debra Cox, Spokane, WA Deluxe Reference Set Ms. Evelyn Swegheimer, Alliance, OH Deluxe Reference Set Ms. Carolyn Current, Ozark, MO NEA Cellular Phone Ms. Adriana Fletes, McKinney, TX NEA Cellular Phone Ms. Suzanne McKaig, Wichita, KS NEA Cellular Phone November 2004: Ms. Brenda Fugett, Middletown, OH $100 Target Gift Card Ms. Angela McCarron, Bogota, NJ $100 Target Gift Card Ms. Lisa Stearns, St. Peters, MO $100 Target Gift Card Mr. Brian Allred, Sellersburg, IN $50 NEA Magazine Service Gift Certificate Mr. Chuck Brickman, Wickliffe, OH $50 NEA Magazine Service Gift Certificate Ms. Mary Butts, Dublin, OH $50 NEA Magazine Service Gift Certificate Mr. James Conyers, Chicago, IL $50 NEA Magazine Service Gift Certificate Ms. Kerry DiGiacinto, West Orange, NJ $50 NEA Magazine Service Gift Certificate Ms. Stephanie Ginder, Fredericksburg, VA $50 NEA Magazine Service Gift Certificate Ms. Altise Harris, Wilson, NC $50 NEA Magazine Service Gift Certificate Ms. Irleny Insuasti, Pembroke Pines, FL $50 NEA Magazine Service Gift Certificate Ms. Robin Richee, Florissant, MO $50 NEA Magazine Service Gift Certificate Ms. Brenda Valentine, Edgard, LA $50 NEA Magazine Service Gift Certificate October 2004: Ms. Kaz M. Wright, New Brunswick, NJ Mutual of Omaha Wild Kingdom School Adventure September 2004: Ms. Christine Bahr, Janesville, WI Teacher Gift Pack Mr. Jacob Barr, Gastonia, NC Teacher Gift Pack Miss Kelly Brown, Cherryville, NC Teacher Gift Pack Mr. Philip Buley, Cornwall, VT Teacher Gift Pack Ms. Edie Butler, Jackson, TN Teacher Gift Pack Ms. Carla Edwards, Plattsburgh, NY Teacher Gift Pack Ms. Susan Farnsworth, Idaho Falls, ID Teacher Gift Pack Ms. Madeline Gottemoeller, New Haven, MO Teacher Gift Pack Ms. Nancy Hackendorn, Newark, DE Teacher Gift Pack Ms. Lisa Ison, Murfreesboro, TN Teacher Gift Pack Ms. Alyson Jones, Kansas City, MO Teacher Gift Pack Dick Knight, Boise, ID Teacher Gift Pack Ms. Amber Konert, Fredonia, NY Teacher Gift Pack Ms. Kathleen Mambourg, Mansfield, OH Teacher Gift Pack Ms. Patti O’Brien, Parlin, NJ Teacher Gift Pack Mr. John Orsulak, Tulsa, OK Teacher Gift Pack Ms. Sally Porter, Colgate, WI Teacher Gift Pack Mr. Ronald Rasp, Alabaster, AL Teacher Gift Pack Mr. Lee Safian, New Milford, NJ Teacher Gift Pack Mr. Robert Schlim, Spokane, WA Teacher Gift Pack Ms. Deedrah Simmons, Reedsville, OH Teacher Gift Pack Mr. David Staat, St. Louis, MO Teacher Gift Pack Ms. Kaarin Schumacher, St. Paul Park, MN Teacher Gift Pack Mr. Jason Van Zee, Springville, IA Teacher Gift Pack Ms. Carol-Jean Whitt, Fairfield, OH Teacher Gift Pack



Birthday Present

Math Forum: Ask Dr. Math FAQ: The Birthday Problem -- Ask Dr. Math: FAQ The Birthday Problem Dr. Math FAQ || Classic Problems || Formulas || Search Dr. Math || Dr. Math Home Suppose you flip a coin and bet that it will come up tails. Since you are equally likely to get heads or tails, the probability of tails is 50%. This means that if you try this bet often, you should win about half the time. What if somebody offered to bet that at least two people in your math class had the same birthday? Would you take the bet? This question is more complicated than flipping a coin, because the chance of finding two people with the same birthday depends on the number of people you ask. If there were only one other person in your math class, you might be surprised to find out that she had the same birthday as you. If there were a pair of people with the same birthday in a class of 366 people, would you still be surprised? How large must a class be to make the probability of finding two people with the same birthday at least 50%? Let's forget about leap year when we solve this problem (no February 29 birthdays!) This way, we can assume that a year is always 365 days long. We'll start by figuring out the probability that two people have the same birthday. The first person can have any birthday. That gives him 365 possible birthdays out of 365 days, so the probability of the first person having the "right" birthday is 365/365, or 100%. The chance that the second person has the same birthday is 1/365. To find the probability that both people have this birthday, we have to multiply their separate probabilities. (365/365) * (1/365) = 1/365, or about 0.27%. Now, what about three people ? The chance of the first and second person sharing a birthday is still 1/365. The first and third person might share a birthday instead. The probability of that is 1/365 as well. But what if the second and third person shared a birthday? And what if all three of them had the same birthday? Things are getting complicated fast. Four or five people would be even messier. Is there a simpler way? To solve the birthday problem, we need to use one of the basic rules of probability: the sum of the probability that an event will happen and the probability that the event won't happen is always 1. (In other words, the chance that anything might or might nothappen is always 100%.) If we can work out the probability that no two people will have the same birthday, we can use this rule to find the probability that two people will share a birthday: P(event happens) + P(event doesn't happen) = 1 P(two people share birthday) + P(no two people share birthday) = 1 P(two people share birthday) = 1 - P(no two people share birthday). So, what is the probability that no two people will share a birthday? Again, the first person can have any birthday. The second person's birthday has to be different. There are 364 different days to choose from, so the chance that two people have different birthdays is 364/365. That leaves 363 birthdays out of 365 open for the third person. To find the probability that both the second person and the third person will have different birthdays, we have to multiply: (365/365) * (364/365) * (363/365) = 132 132/133 225, which is about 99.18%. If we want to know the probability that four people will all have different birthdays, we multiply again: (364/365) * (363/365) * (362/365) = 47 831 784/ 48 627 125, or about 98.36%. We can keep on going the same way as long as we want. A formula for the probability that n people have different birthdays is ((365-1)/365) * ((365-2)/365) * ((365-3)/365) * . . . * ((365-n+1)/365). If you know permutation notation, you can write this formula as (365_P_n)/(365^n). That's the same as 365! / ((365-n)! * 365^n). We've made some progress, but we still haven't answered the original question: how large must a class be to make the probability of finding two people with the same birthday at least 50%? We know that the probability of finding at least two people with the same birthday is 1 minus the probability that everybody has a different birthday, and we know how to find the probability that everybody has a different birthday for any number of people. The easiest way to find the right class size is to use a calculator to try different numbers in the formula. It turns out that the smallest class where the chance of finding two people with the same birthday is more than 50% is... a class of 23 people . (The probability is about 50.73%.) From the Dr. Math archives: Probability Theory: Coincidental Birthday Probability of the Same Birthday within a Group Birthday Probabilities Three Share a Birthday The Birthday Problem; Queuing at a Bank Birthday Probability, Class of 25 One Person of Seven Born on Monday Odds of Left-Handedness in a Group From the Web: The Birthday Problem: A short lesson in probability , George Reese A Java applet that you can use to test different class sizes (it works better with small classes) and graphs of the probability for different numbers of people. The Law of Small Errors , Keith Devlin The birthday problem, and related questions - what's the probability that someone will have your birthday? Birthday Surprises, Ivars Peterson Birthday Problem, Eric Weisstein's World of Mathematics Coincidence, Alexander Bogomolny How to Read Mathematics, Shai Simonson and Fernando Gouveau This article uses an explanation of the birthday problem as an example. An Introduction to Mathematica and the "Birthday Problem," Louie Beuschlein For a general review of probability: Probability, Dr. Math FAQ Probability in the Real World, Dr. Math FAQ - Ursula Whitcher, for the Math Forum Submit your ownquestion to Dr. Math [ Privacy Policy ] [ Terms of Use ] Math Forum Home || Math Library || Quick Reference || Math Forum Search Ask Dr. Math ® © 1994-2005 The Math Forum http://mathforum.org/dr.math/




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