₦airaland Forum

naturalwaves:
I think I get what you are trying to say now

Modified :
....... but you must realize that what you're asking is impossible to determine without towing the probability lane. The best we can do is to calculate the probability of an unshared birthda.y and then tweak it. Take for instance if the probability of an unshared birthday is 1/5. It means that for every 5 people, 1 won't share birthday and since we are looking at a total of 100 students, you will get 20 students in all as what you desire.

Unfortunately, in this case the probability is marginal and the number you require is about 3 in a 10 million samples or something like that.

Cc: jackpot, Mr. Shape

naturalwaves:
A random sample can also give you 0 people without birthday mates in a size of 100.

Mrshape:
3 out of 10000000 people

naturalwaves:
Thanks for the clarification.
This is my points exactly.
Based on this example, 15 people have shared birthdays but the number of shared birthdays is just 7. That's the difference I was talking about.

naturalwaves:
The question requires.... Number of unshared birthdays from the group and not number of students with an unshared birthday.

naturalwaves:
Thanks Mr. Shape for talking about the birthday paradox. I had to quickly read on it on wiki and then I was lucky to get this video. Please, let us take time to watch it. The dude in the video even used 100 people to illustrate from time stamp 5:33 Please, try and watch.
CC: Martinez39s, jackpot.


https://www.youtube.com/watch?v=ofTb57aZHZs

Martinez39s:


Your question doesn't make a shred of sense, or it is incomplete. If I have 100 students in a class, I need to know their birthdays before I can know the expected number of students that have no birthday mates because there are 365¹⁰⁰ possible ways a group of 100 students can be assigned birthdays.

Let's pick four possibilities with a class of 100:
1) If no one has a birthday mate, the expected number of students with no birthday mates is 100.
2) If only two students share the birthday and others don't have birthday mates, the expected number of students with no birthday mate is 98.
3) if three students share the same birthdays, two students share another birthday, and the rest have no birthday mates, the expected number of students with no birthday mates is 95.
4) If all students share the same birthday, the expected number of students with no birthday mates is 0.

So you see, you need to know the birthdays of each student to know the expected number of students with no birthday mates.

Edit: I could list many more possibilities. In fact, I can list a possibility to arrive at any expected number (from 0 to 100) of students with no birthday mates.

naturalwaves:
See what you wrote now Jackpot
You initially said 15 have shared birthdays.
Now, 6 pairs =12
Hiw did they share different birthdays again. I think you didn't understand my question.

I actually meant that, for the number that have their birthdays shared, what is the distribution like? E. G if 15 people share birthdays, 4 can be April 30 and the remaining 11 can be May 1. That's the sort of info I requested.

Mrshape:
The question how many birth dates are shared can't be answers with does information, it is a lot of possibilities

naturalwaves:
That means that 15 players share birthdays but only 12 birthdays are shared.

naturalwaves:
You've not answered my question. From the 15 people with shared birthdays, how many birthdays are shared
15 people can share just 1 birthday.

Martinez39s:
I said it, but others were telling me I need probability. Naturalwaves found the probability of selecting a group of 100 in which at least two share a birthday assuming we have 365 days in a year. His solution doesn't make sense because number of birthdays can't be a decimal, neither can it be a probability.

Nevertheless, I am yet to understand your question.
•• If you are looking for the number of possible ways in which no two people in a group of 100 share the same birthday (assuming we have 365 days in a year) then your answer is

•• If you seek the number of possible ways in which at least two people in a group of 100 share the same birthday (assuming we have 365 days in a year) then your answer is

I repeat, I still don't understand your question except I presume what you could possibly mean.

Mrshape:
He did the correct thing sis

naturalwaves:
You should also be careful not to confuse "unshared birthdays" with "number of students with unshared birthdays" The question requires the former.

naturalwaves:
No, my answer says it is almost impossible to see an unshared birthday which means that the probability of having a shared birthday is higher.

naturalwaves:
The question is not as basic as you see.

naturalwaves:
There is a large difference between 15 students sharing birthday and number of shared birthdays 15 students can share just 1 birthday. 15 students can share 3 birthdays etc etc .

naturalwaves:
How many birth days are shared in the document?

naturalwaves:
Multiply the answer by 100 to get your answer. That's a negligible figure as it is far far less than 1.

jackpot:
Maybe we should try to simulate it using real data. This is the [url]https://img.fifa.com/image/upload/hzfqyndmnqazczvc5xdb.pdf [/url]for 2018 fifa world cup data with the players birthdays.

We can decide to pick any consecutive 100 players. Either from top, middle or bottom.

Let's go!

naturalwaves:
SOLUTION
We need to take a step by step approach in order to get this.
Since 29th of February is not included, this means that we are dealing with just 365 days.
Having said that, what is the expected probability of the 1st student? That is going to be 365/365 days
What is the expected probability of the 2nd student? 364/365
Third? 363/365
Fourth....362/365
Fifth..................361/365
This is how it will go on till we get to the 100th student which will be 266/365 but we cannot do this as it will be too cumbersome.
Now, let us sort the denominator. Every component in the denominator is 365 i.e 365/365 * 364/365 * 363/365................*266/365.
This means that the denominator will be 365¹⁰⁰.

How about the numerator? We have 365! but we are not going to be needing the whole of it. We only need the area from 365 to 266. How do we get this?
All we need do is to divide 365! by 265!.

Thus, the desired solution to the problem is;
(365!) / (265! * 365¹⁰⁰).

Modified:
So, I have gotten a calculator,
(365!) / (265! * 365¹⁰⁰). = 3.0725 * 10^-7 approximately.
That is around 0.00000030725 approx.
As you can see, the probability is very slim as it is almost unlikely that out of 100 students, none will share a birthday which makes a lot of sense.

cc: Jackpot, Dejt4u, Martinez39s, Mrshape, Richiez, mathefaro,ifada123, Goalnaldo

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ifada123:
Let's go.

Probability of first person number having birth day with another is 364/365
Second 363/365
Third 362/365
......... 265/365
So
Prob of not have same of 100 persons will be =

(First ) and (second ) and (third)and....×(100th no)
=(First not) x (second not) x (third not)x.....(100th not)
(364/365)×(363/365)×(362/365).......x(265/360)
=(1/365) x 364 P100
=

Almost everyone will share same birth day

ifada123:
Is require critical thinking o

Martinez39s:
Do you actually mean the number of possible ways in which the students can have their birthdays without any two sharing the same birthday? If that is what you mean, your answer is

dejt4u:
N =100
P = 1 - (1/365) = 364/365

Expected value = N×P = 100 × (364/365) = 99.73

Mrshape:
Thanks that's were missed it.
I was supposed to get 20° as the answer

Mrshape:
Use sin rule

Deltayankeeboi:
Sweet in the middle kind of wife.

cococandy:
Two options

Get a boyfriend that looks similar to your husband and have a baby for him. Make your husband pay for the kid until the kid is grown and then tell him the truth.

Or.....you’re only 26, you have the world ahead of you. You can dump him and find someone else.

That’s on the off chance that you’re not another troll account created for sensational stories.

No hypocrite should quote me to talk nonsense please. Thanks.