Probability (1 Viewer)

[StGeorgeDragons]

u have 2 people in a room, what is the probability they have there birthday on the same day?

 

[haboozin]

u have 2 people in a room, what is the probability they have there birthday on the same day?

1 - 365p2/365^2

= 2.739726027 x 10^-3
= 1/365
if theres 23 people theres 50% chance
and 45 people.... 90%

 

[Stefano]

1 - 365p2/365^2

= 2.739726027 x 10^-3
= 1/365
if theres 23 people theres 50% chance
and 45 people.... 90%

What's your reasoning haboozin ?

the "1-..." suggests you are working out the probability that they do NOT share the same birthday.

Could you explain what 365P2 represents etc ??

Btw; Should we be using 365.25 as a better approximation ?


||| It's interesting how high the probability becomes as the amount of people increases! This is due to the fact that its dependant on the amount of combination and, ofcourse, that the amount of days never changes Just imagine, World Cup Soccer Final... in that stadium - how many people share the same birthday and what are the chances!?||||

 

[StGeorgeDragons]

sry i dont understand. can u please explain wat u have done

 

[kami]

I'm probably going to sound incredibly stupid for this but...wouldn't it just be (1/365)²
You *could* also do it via Binomial Distrubition...but what would be the point?

 

[Stefano]

coderer:

365^2 = TOTAL amount of permutations (birthdays)
In otherwords, one person's birthday falls on 1 of the 365 days right? So, if you have 2 people then it must be (1/365) x (1/365) = 1 of the 365^2 days. Capron ?

36P2 = Not sure

The reason he is subtracting it all from 1 is because he is working out the opposite to the question (to save time because he is lazy ) and we know that P(e) + P'(e) = 1

Hope that helps ?

 

[Stefano]

I'm probably going to sound incredibly stupid for this but...wouldn't it just be (1/365)²
You *could* also do it via Binomial Distrubition...but what would be the point?

Kami, I'll try to eplain this without mathematics:

If you put it in the form (1/365)^n then the probability gets SMALLER as n INCREASES.

This means that the more people you have the less chance they will share the same birthday. However, this defies logic! If you had say, 100,000,000 people then SURELY some of them share the same birthday! Likewise, if you only have 2 then it's highly unlikely.

 

[KFunk]

Correct me if I'm wrong but wouldn't the easiest way be:

Person 1 has a birthday
The likelihood that person 2 has this certain birthday = 1/365

∴ P(same B'day) = 1/365

 

[...]

Correct me if I'm wrong but wouldn't the easiest way be:

Person 1 has a birthday
The likelihood that person 2 has this certain birthday = 1/365

∴ P(same B'day) = 1/365

leap year??

 

[KFunk]

Most questions seem to ignore leap years but add it in if you want.

 

[Stefano]

Correct me if I'm wrong but wouldn't the easiest way be:

Person 1 has a birthday
The likelihood that person 2 has this certain birthday = 1/365

∴ P(same B'day) = 1/365

No correction necessary. You are exactly correct.
Good reasoning; true math.

I just wanted to show the theory because it won't always just be 2 people

PS> Ignore leap years. It's just 3u.

 

[StGeorgeDragons]

1 - 365p2/365^2

= 2.739726027 x 10^-3
= 1/365
if theres 23 people theres 50% chance
and 45 people.... 90%

i dnt understand. how do u get 90% if theres 45 people.

 

[Stefano]

i dnt understand. how do u get 90% if theres 45 people.

Replace the '2' with '45' !

1 - (365P45)/365^45
=(approx) 0.9

Note: These numbers are probably too big for your calculator. So just take our word for it

 

[...]

Most questions seem to ignore leap years but add it in if you want.

thats the problem
how do you take leap year into consideration

 

[insert-username]

You usually ignore the leap year, if I recall correctly. To factor it in, you'd have to multiply in the chance that someone is born in a leap year on February 29 - 1/4 x 1/366, I think. NNot too sure on that one.

365P2 is permutation notation - 365 total events, and you want the probability of selecting 2 of those events ignoring order. Or, "how many ways can I select 2 days from 365 ignoring their order?"


I_F

 

[StGeorgeDragons]

how did u get that formula? why do u do 1 minus. do u devide by 365 power of n cause its the total amount of events?

 

[insert-username]

That probability of an event that will always happen is 1. Similarily, the probability of an event will never happen is 0. Thus, you subtract from 1 to find your probability. 365P2 the number of permutations, and is a shorthand way of writing:

n!/(n-r)!

Where n! is the total amount of events, and r is the amount you're looking for. So 365P2 is

365!/(365-2)!

You divide by 365 because probability is worked out in fractions, with the probability of 1/1 being certain and 0/1 impossible. 1/365 is the probability of 1 chance out of 365.


I_F

 

[KFunk]

thats the problem
how do you take leap year into consideration

Hmm, if person 1 was born on a leap day then the probability that they share a birthday is 1/1461. If person 1 wasn't born on such a day then the probability is 4/1461.

To factor both in then you need to find P(both born on a leap day)and P(both are not born on a leap day). Giving you 1460²/1461² and 1/1461² respectively. We know that if they are both born on a normal day of the year then P(same B'day) = 1/365.

I havn't entirely thought this through but I think that if you factor in leap years then:

P(same B'day) = (4x1460 + 1)/1461² = 2.736 x 10^-3

 

[haboozin]

sry i dont understand. can u please explain wat u have done

hey,
if u dont understand, you dont even need it for this question

the first person can have any birthday
so his probability is 1, now, second person has 1/365 chance of having the same birthday as the first person.

Leap year is ignored.

Anyways:
if this was say, more than 2 people you would be better off doing it 1 - probability of different birthday

that is

365/365 x 364/365 x 363/365 .... 1/365

equating them: 365pn/365^n

now 1 - 365pn/365^n

but u dont need this for this question anyways.

I think it was mentioned in the cambridge book.

 

[currysauce]

P (Same bday) = 1/365