Question

PLZ HELP!! Last Question!

Answer 1

What is your question?

Answer 2

can u see the attachment?

Answer 3

I'm sorry.......I can't help u with ur question :(

Answer 4

ok thx :(

Answer 5

this one is messier to set up.

Answer 6

well u could always just give me the answers. ;)

Answer 7

i don't know the answers!! :) Wish it was that easy!

Answer 8

well dang :(

Answer 9

the first person can have 365 different birthdays
Person 2 can have 364 different birthdays if she doesn't share with Person 1
Person 3 can have 363 different birthdays if she doesn't share with Person 1 or 2

Answer 10

ok?

Answer 11

So, in terms of probability, the cases that "win" (like the lottery ticket) are those where all 13 people have different birthdays.

There are 365 x 364 x 363 x 362 x 361 x 360 x 359 x 358 x 357 x 356 x 355 x 354 x 353 different winning combinations where everyone has DIFFERENT birthdays.

Answer 12

A probability is the # of winners / total # of outcomes

Total # of outcomes is like:
Each person can have any one of 365 days as a birthday.
Total number of birthday combinations, including ones where people may share, is 365 x 365 x... thirteen times... 365^13

Answer 13

ok i'm doing it wrong. it keeps comeing out as a scientific notation

Answer 14

you have to do a shortcut, I think... I'm trying to remember.

You can't do all that multiplying... too big... need to cancel terms first.

Answer 15

i'm starting to not care if i get it right or not. lol my brain is tired.

Answer 16

ugh. same here, and it's not my hw.

0.8056 appears to be the 4-digit approximation of the probability. I can't remember how to set it up to avoid the giant numbers...

I found this site that would calculate it, but I'm not totally positive I didn't mess it up.

http://www.wolframalpha.com/input/?i=%28365+factorial%2F352+factorial%29%2F%28365^13%29

Answer 17

I am going to have to take off... sorry I couldn't help more on this one.

Good luck!

Answer 18

ok thx!! now to do b!!