Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

xlegendx Group Title

LEGEN...waitforit...DARIDDLE: How many people should be in a room so that there is a 99% probability that two or more people share the same birthday? Hint: whoever gets it right with solutions gets a medal

  • one year ago
  • one year ago

  • This Question is Closed
  1. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    the dreaded riddles are back!

    • one year ago
  2. xlegendx Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    these ones are legendary. i don't make puny riddles like the original

    • one year ago
  3. xlegendx Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so @hartnn do you know the answer?

    • one year ago
  4. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    haven't tried yet...i would always prefer helping than satisfying my own thirst of solving puzzles...i'll try that in free time :)

    • one year ago
  5. xlegendx Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    helping huh. haven't tried that in a while

    • one year ago
  6. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    i try these problems by taking small numbers first let there be just 2 people in that room then the probability "that two or more people share the same birthday" will be 1- probability that BOTH will not have same birthday (assuming 366 days a year) so, \(\large 1- \dfrac{366}{366}\times \dfrac{365}{366}\) now extending this for 'n' people \(\large 0.99 =1- \dfrac{\dfrac{366!}{(366-n)!}}{366^n}\) am i on right path ? (wondering how the hell can i solve that for n :O O.o)

    • one year ago
  7. xlegendx Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    lol. am not saying if you're on the right path or not. just tell me the answer and i'll tell you if it's right. that's how riddles work

    • one year ago
  8. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    rounding to nearest integer i am getting 55 people (infact 55 or more!)

    • one year ago
  9. xlegendx Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ooooh so close

    • one year ago
  10. xlegendx Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    but no

    • one year ago
  11. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    i had my chance, i'll let others try....

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.