dejt4u:
you can your expected values from probability.. It is even simpler in this case since our N is 100

dejt4u:
but in reality, this is not valid

Martinez39s:
Jackpot never mentioned anything about probability. How did probability get involved in the solution?

Mechanics96:


Exactly what I have solved here, *just added that the probability of having at least a pair of students whose birthday is the same will be:
1-0.0000003073 which is 0.99999... And that's approximately 1.

So if we pick out a pair of students at random assuming that they are sure to share birthdays (probably 1) then we have 98 students left, out of which there may still be some pairs of same birthdays. There's got to be how to get a specific number of students who are likely not to share birthdays... Maybe someone can help figure this out.

Martinez39s:
Jackpot never mentioned anything about probability. How did probability get involved in the solution?

naturalwaves:

Lmao. She used term expected number in a group of 100 which points towards probability. If you need to get the number out of 100,just multiply the answer by 100. E. G if you have (1/4) =0.25
as a probability which means 1 out of every 4. If you now want to get the 1,you will need to multiply 4 by 0.25.So,multiply 100 by the answer to get the figure. You won't still get up to even 1. What does that say? It means that the probability that no 2 persons will share same birthday out of the 100 is negligible.

Mechanics96:


Exactly what I have solved here, *just added that the probability of having at least a pair of students whose birthday is the same will be:
1-0.0000003073 which is 0.99999... And that's approximately 1.

So if we pick out a pair of students at random assuming that they are sure to share birthdays (probably 1) then we have 98 students left, out of which there may still be some pairs of same birthdays. There's got to be how to get a specific number of students who are likely not to share birthdays... Maybe someone can help figure this out.

dejt4u:
always proud of you. You're a good teacher. Keep it up bro!

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)^100 x 364 P100
=

Almost everyone will share same birth day

naturalwaves:


The last number for the 100th was supposed to be 266 and not 265. Your solution is similar to mine just that you were rushing when solving.

You were supposed to compute as (365!)/(265!*365¹⁰⁰).

naturalwaves:


If the probability of having 2 students with the same birthday like you said is almost 1 which I agree with. THEN, the probability of not having 2 same birthdays will be extremely close to 0.

Moreover,the fact that there are 100 physical students does not mean that to get the figure from the group, you must have a digit or something. Your probability can even tend far far towards 0. Check my solution on the previous page.

Mechanics96:
.


I saw your solution, I solved it same way too.

Thanks for adding this �

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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jackpot:
I think that you are solving a different question

The question says expected number of unshared birthdays among 100 students?

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!

jackpot:



Okay, for the first 100 players in the PDF whose link is shown, I found out that
15 share birthdays while 85 didn't share birthdays.

However, this gives just an idea of the problem but is not the answer at all

naturalwaves:

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

jackpot:



I think that the answer you got is the probability of 100 persons having distinct birthdays.


I don't think E=NP should work here. If you multiply by 100, it's still a negligible number and it will tend to invalidate your earlier probability (because I think that your probability answer is saying that it is almost impossible not to see a shared birthday among 100 persons). Check again.

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?

jackpot:
Sure.

From that data, 85 have distinct birthdays but 15 must have at least a birthday mate. Note that as you noticed, this is different from saying that 15 share the same birthday.
A lot. I just considered the first 100 so that I will not deviate from the question.

jackpot:



Okay, for the first 100 players in the PDF whose link is shown, I found out that
15 share birthdays while 85 didn't share birthdays.

However, this gives just an idea of the problem but is not the answer at all

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:

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

jackpot:
I think that you are solving a different question

The question says expected number of unshared birthdays among 100 students?

jackpot:
I think that you are solving a different question

The question says expected number of unshared birthdays among 100 students?

³⁶⁵P₂₆₅ = 365!/265!

365¹⁰⁰ – (³⁶⁵P₂₆₅) = 365¹⁰⁰ – 365!/265!

Mrshape:

He did the correct thing sis

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.

jackpot:
the birthdays belong to somebody na

What's the difference?

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.

jackpot:

Among the 15 players, 6 pairs of players shared 6 different birthdays while 3 players shared the same birthday.

So, 6×2+3=15.

The rest 85 players do not have birthday mates.

jackpot:
His answer is correct but I think it is not the direct answer to the question I posed.