Question

What is the probability that at least 2 people have the same birth month in a group of 6 people?

Answer 1

what grade is this?

Answer 2

what grade is this for

Answer 3

well its algebra 2

Answer 4

ok on it

Answer 5

thx

Answer 6

ok got it

Answer 7

do u want the answer or explained?

Answer 8

explained pls

Answer 9

ok 1 sec brb

Answer 10

so
lets begin

Answer 11

to start this problem, it is helpful to calculate the exact opposite of what u said, in other words, calculate the porbability that in a group of 6, NO ONE will have same birth month

Answer 12

okay

Answer 13

so thats 1/12

Answer 14

no, not exactly

Answer 15

or is it (1/12)^6

Answer 16

it is equal to \[\frac{ \prod_{7}^{12} }{ 12^{6} }\]

Answer 17

what is that thing on top????

Answer 18

this may look intimidating, but it is simply (12/12)(11/12)(10/12)(9/12)(8/12)(7/12)

Answer 19

oh is that a permutation on top?

Answer 20

okay yea that makes sense

Answer 21

the reason for which this is correct is that the first term is the probability that the first person has the same birthday as himself (12/12). The second term is the probability that the SECOND person has a DIFFERENt birthday than the first. The THIRD term is the prob that the THIRD person has a dif birthday from the first and second ppl and so on down the sixth person

Answer 22

multiplying all these terms gives the summative probability that in a group of 6, NO ONE will have same birth month

Answer 23

mhm or u could just do the permutation of the whole thing and divide it by the extra :D

Answer 24

nah thats not a permutation

Answer 25

its a product sign. it means multiply all the terms from 7 to 12

Answer 26

is it a permutation on top? xD

Answer 27

okay nvm I'll just write it all out.

Answer 28

heres the thing, permutations wont help here because its strictly conditional

Answer 29

ok

Answer 30

that is NOT a permutation, it is simply a condensed version of the product (12/12)(11/12)(10/12)(9/12)(8/12)(7/12)

Answer 31

ight

Answer 32

do u understand why i am multiplying these terms and what they mean?

Answer 33

yup

Answer 34

ok cool now from here

Answer 35

from here, i need u to understand that the probablitiy that in a group of 6, NO ONE will have same birth month, PLUS the probability that in a group of 6, AT LEAST 2 PPL will have same birth month, is equal to EXACtly 1. Do u understand why this is?

Answer 36

why is that equal to 1?

Answer 37

yes? no?

Answer 38

ok, that is equal to 1 because it includes ALL THE POSSIBLITIES. look at it this way, either i have the same birthmonth as you, or i dont!

Answer 39

get it now? brb btw

Answer 40

oh okay i get it sort of..

Answer 41

alrite so the prob i dont have the same bday as anyone in my group of 6, is P(d). the prob that i have the same b day w at least another person in my group of 6, is P(s). Since this encompasses EVERYTHING, ALL POSSIBLITIES, then P(d) + P(s) = 1, or 100%. get itnow?

Answer 42

yeah i got it now

Answer 43

alrite so can you calculate that P(d) for me? use the (12/12)(11/12)...(7/12) i gave u before

Answer 44

helooooo thatonegirlll

Answer 45

.2228 @rizags sorry went to eat

Answer 46

alrite. so subtract that from 1, then multiply by 100 and you get your answer in a percent

Answer 47

medal plz :)

Answer 48

Thanks so much!! @rizags